The Pythagorean Proposition

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The Pythagorean Proposition

ELISHA S. LOOMIS

Photo 1935

The Pythagorean Proposition

Its Demonstrations Analyzed and Classified

And

Bibliography of Sources

For Data of

The Four Kinds of “Proofs”

SECOND EDITION

1940

By

Elisha S. Loomis, Ph. D., LL. B.

Professor Emeritus of Mathematics
Baldwin-Wallace College

Professor Mathematics, Baldwin University, 1885-1895, Head
of Mathematics Department, West High School,
Cleveland, Ohio, 1895-1923
and

Author of “Original Investigation or How to
Attack an Exercise in Geometry*'

(Ginn & Co., 1901)

Copyright, 1940, by
ELISHA S. LOOMIS

(Made and printed in the United States of America)

EDWARDS BROTHERS, INC.
LITHOPRINTERS
ANN ARBOR, MICHIGAN
1940

Dedicated

to

DR. OSCAR L. DUSTHEIMER, A.M., Ph.D.
Professor of Mathematics and Astronomy
Baldwin-hallace College
Berea, Ohio

"In Mathematics the man who is ig-
norant of what Pythagoras said in Croton in
500 B.C. about the square on the longest
side of a right-angled triangle, or who for-
gets what someone in Czechoslovakia proved
last week about inequalities, is likely to
be lost. The whole terrific mass of well-
established Mathematics, from the ancient
Babylonians to the modern Japanese, is as
good today as it ever was. "

E. T. Bell, Ph.D., 1931

vl

FOREWORD

According to Hume, (England's thinker vho
Interrupted Kant's "dogmatic slumbers"), arguments
may be divided into: (a) demonstrations; (b) proofs;
(c) probabilities.

By a demonstration , (demonstro, to cause to
see), we mean a reasoning consisting of one or more
catagorlcal propositions "by which some proposition
brought into question is shown to be contained in
some other proposition assumed, whose truth and cer-
tainty being evident and acknowledged, the proposi-
tion in question must also be admitted certain. The
result is science, knowledge, certainty." The knowl-
edge which demonstration gives is fixed and unalter-
able. It denotes necessary consequence, and is
synonymous with proof from first principles.

PJ^oof 1 (probo, to make credible, to demon-
strate), we mean 'such an argument from experience
as leaves no room for doubt or opposition' ; that is,
evidence confirmatory of a proposition, and adequate
to establish it.

The object of this work is to present to the
future investigator, simply and concisely, what is
known relative to the so-called Pythagorean Proposi-
tion, (known as the 47th proposition of Euclid and as
the "Carpenter's Theorem”), and to set forth certain
established facts concerning the algebraic and geo-
metric proofs and the geometric figures pertaining
thereto.

It establishes that;

First , that there are but four kinds of demon-
strations for the Pythagorean proposition, viz.:

I. Those based upon Linear Relations (im-
plying the Time Concept) — the Algebraic Proofs.

vll

vlll

THE PYTHAGOREAN PROPOSITION

II. Those based upon Comparison of Areas
(Implying the Space Concept ) --the *Xjeometric Proofs.

III. Those based upon Vector Operation (im-
plying the Direction Concept) — the Quaternlonic
Proofs.

rv. Those based upon Mass and Velocity (im-
plying the Force Concept) — the Dynamic Proofs.

Second , that the number of Algebraic proofs
is limitless.

Third , that there are only ten types of geo-
metric figures from which a Geometric Proof can be
deduced.

This third fact is not mentioned nor implied
by any work consulted by the author of this treatise,
but which, once established, becomes the basis for
the classification of all possible geometric proofs.

Fourth , that the number of geometric proofs
is limitless.

Fifth , that no trigonometric proof is possi-
ble.

By consulting the Table of Contents any in-
vestigator can determine in what field his proof
falls, and then, by reference to the text, he can
find out wherein it differs from what has already
been established.

With the hope that this simple exposition of
this historically renowned and mathematically funda-
mental proposition, without which the science of Trig-
onometry and all that it implies would be Impossible,
may Interest many minds and prove helpful and sugges-
tive to the student, the teacher and the future orig-
inal investigator, to each and to all who are seeking
more light, the author, sends it forth.

CONTENTS

Figures Page

Foreword vll

Portraits xl

Acknowledgments xlll

Abbreviations and Contractions xv

The Pythagorean Proposition 3

Brief Biographical Information 7

Supplementary Historical Data 11

An Arlthmetlco-Algebralc Point of View 17

Rules for Finding Integral Values for a, b and h . 19

Methods of Proof--4 Methods 22

I. Algebraic Proofs Through Linear Relations . , 23

A. Similar Right Triangles — several thousand proofs

possible 1-35 23

B. The Mean Proportional Principle 36 55 51

C. Through the Use of the Circle 54- 85 60

(l) Through the Use of One Circle ... 54- 85 6 O

( 1 ) The Method of Chords 54- 68 6 l

( 2 ) The Method by Secants ..... 69 - 76 68

(3) The Method by Tangents 77-85 7^

(II) Through the Use of Two Circles ... 86-87 8 O

D. Through the Ratio of Areas ....... 88 - 92 83

E. Through the Theory of Limits 93- 94 86

F. Algebraic -Geometric Complex 95- 99 88

G. Algebraic -Geometric Proofs Through Similar

Polygons, Not Squares IOO-IO 3 91

II. Geometric Proofs- -10 Types 97

A-Type. All three eq *8 const 'd exterior , 104-171 98

B-Type. The h-square const *d Interior . . 172-193 l44

C-T^pe. The b-sqiiare const ’d Interior . . 194-209 158

D-Type. The a-square const ’d Interior . . 210-216 I 65

E-Type. The h- and b-sq's const *d Interior 217-224 I 69

T^Type, The h- and a-sq's const 'd Interior 225-228 174

G-iype. The a- and b-sq's const ’d Interior 229-247 I 76

H-l^e. All three eq*e const *d Interior . 248-255 l85

lx

X

THE PYTHAGOREAN PROPOSITION

Flgiores Page

One or more squares translated, giv-

ing 7 classes covering 19 different cases • . . • 189

(1) Four cases 256-260 191

(2) Four cases 261-276 193

(5) Four cases 277-288 201

(h) Two cases 289-290 206

(5) Two cases 291-293 208

(6) Two cases 29^-305 209

(7) One case 306 215

J -lype . One or more of the squares not

graphically represented — Two sub -types 2l6

(a) Proof derived through a square, giv-
ing 7 classes covering 45 distinct

cases 216

(1) Twelve cases, but 2 given .... 307-309 2l8

(2) Twelve cases, but 1 given .... 310-311 219

(3) Twelve cases, none given .... 0

(4) Three cases, 312-313 221

(5) Three cases, none given 0

(6) Three cases, all 3 given • • • . 314-328 222

(B) Proof based upon a triangle through

calculations and comparisons of 230

equivalent areas 329-346 230

Why No Trigonometric, Analytic Geometry

Nor Calculus Proof Possible 244

III. Quaternlonlc Proofs 547-550 246

IV. Dynamic Proofs 351-353 248

A Pythagorean Curiosity 354 252

Pythagorean Magic Squares 355-359 254

Addenda 360-366 258

Bibliography 271

Testimonials 277

Index 281

PORTRAITS

1. Loomis, Elisha S

2. Copernicus . .

5. Descartes . . .

4. Euclid ....

5. Galileo ....

6. Gauss

7. Leibniz ....

8. Lobachevsky . .

9. Napier ....

10. Newton ....

11. Pythagoras . . ,

12. Sylvester . . . ,

. . Frontispiece
facing page 88

II If 244

" 118

" ” 188

II • If

” ” 58

II It 210

" M 144

" " 168

It tt Q

” ” 266

xl

ACKNOWLEDGMENTS

Every man builds upon his predecessors.

My predecessors made this work possible, and
may those who make further investigations relative
to this renowned proposition do better than their
predecessors have done.

The author herewith expresses his obliga-
tions :

To the many who have preceded him in this
field, and whose text and proof he has acknowledged
herein on the page where such proof is found;

To those who, upon request, courteously grant-
ed him permission to make use of such proof, or refer
to the same;

To the following Journals and Magazines whose
owners so kindly extended to him permission to use
proofs found therein, viz.:

The American Mathematical Monthly;

Heath’s Mathematical Monographs;

The Journal of Education;

The Mathematical Magazine;

The School Visitor;

The Scientific American Supplement;

Science and Mathematics; and

Science .

To Theodore H. Johnston, Ph.D. , formerly
Principal of the West High School, Cleveland, Ohio,
for his valuable stylistic suggestions after reading
the original manuscript in 1907 *

To Professor Oscar Lee Dusthelmer, Professor
of Mathematics and Astronomy, Baldwin-Wallace College,
Berea, Ohio, for his professional assistance and ad-
vice; and to David P. Simpson, 55°, former Principal
of West High School, Cleveland, Ohio, for his brother-
ly encouragement, and helpful suggestions, 1926.

xiil

xlv

THE PYTHAGOREAN PROPOSITION

To Dr. Jehuthlel Ginsburg, publisher of
Scrlpta Mathematlca, New York City, for the right to
reproduce the photo plates of ten of his "Portraits
of Eminent Mathematicians."

To Elatus G. Loomis for his assistance in
drawing the 366 figures which appear in this Second
Edition.

And to "The Masters and Wardens Association
of The 22nd Masonic District of the Most Worshipful
Grand Lodge of Free and Accepted Masons of Ohio,"
owner of the Copyright granted to it in 1927 ^ f'or its
generous permission to publish this Second Edition
of The Pythagorean Proposition, the author agreeing
that a complimentary copy of it shall be sent to the
known Mathematical Libraries of the World, for pri-
vate research work, and also to such Masonic Bodies
as it shall select. (April 27, 19^0)

ABBREVIATIONS AND CONTRACTIONS

Am. Math. Mo. = The American Mathematical Monthly,

100 proofs, 189 ^.

a- square = square upon the shorter leg.
b- square = " ” ” longer leg.

Colbrun = Arthur R. Colbrun, LL.M., Dist. of Columbia
Bar.

const. = construct,
const *(i = constructed,
cos = cosine.

Dem. = demonstrated, or demonstration.

Edw. Geom. = Edward* s Elements of Geometry, 1895.
eq. = equation.
eq*s = equations.

Pig. or fig. = figure.

Pourrey = E. Pourrey*s Curiosities Geometrlques .

Heath = Heath *s Mathematical Monographs, 19OO,

Parts I and II--26 proofs,
h- square = square upon the hypotenuse.

Jour. Ed'n = Journal of Education.

Legendre = Davies Legendre, Geometry, I 858 ,

Math. = mathematics

Math. Mo. = Mathematical Monthly, I 858 - 9 .

Mo. = Monthly.

No. or no. = number.

01ney*s Geom. = 01ney*s Elements of Geometry, Uni-
versity Edition.
outw*ly = outwardly,
par. = parallel,
paral. = parallelogram,
perp. = perpendicular,
p. = page,
pt. = point,
quad. = quadrilateral.
resp*y = respectively.

XV

xvi

THE PYTHAGOREAN PROPOSITION

Richardson = John M. Richardson- -28 proofs,
rt . = right .

rt. trl. = right triangle,
rect. = rectangle.

Scl. Am, Supt. = Scientific American Supplement,

1910 , Vol. 70 .
sec = secant,
sin = sine,
sq. = square.
sq*s = squares,
tang = tangent.

/. = therefore,
tri, = triangle.
tri*s = triangles,
trap. = trapezoid.

V. or V. = volume.

Versluys = Zea en Negentlc (96) Beweljzen Voor Het
Theorems Van Pythagoras, by J. Versluys, 1914.
Nipper = Jury Nipper *3 ”46 Bevelse der Pythagor-
aischen Lehrsatzes,” I 88 O.

HE , or any like symbol = the square of, or upon, the
line HE, or like symbol.

AC

ACIAP, or like symbol = AC + AP, or — • See proof 17*

dfuXeiv

E

GOD GEOMETRIZES

THE PYTHAGOREAN PROPOSITION

This celebrated proposition is one of the
most Important theorems in the whole realm of geome-
try and is known in history as the 47th proposition,
that being its nimiber in the first book of Euclid’s
Elements •

It is also (erroneously) sometimes called the
Pons Aslnorum. Although the practical application
of this theorem was known long before the time of
Pythagoras he, doubtless, generalized it from an Egyp-
tian rule of thiamb (3^ + 4^ = 5^) and first demon-
strated it about 540 B.C., from which fact it is gen-
erally known as the Pythagorean Proposition, This
famous theorem has always been a favorite with geo-
metricians.

(The statement that Pythagoras was the in-
ventor of the 47th proposition of Euclid has been de-
nied by many students of the subject.)

Many purely geometric demonstrations of this
famous theorem are accessible to the teacher, as well
as an unlimited number of proofs based upon the al-
gebraic method of geometric investigation. Also
quaternions and dynamics furnish a few proofs.

No doubt many other proofs than these now
known will be resolved by future investigators, for
the possibilities of the algebraic and geometric re-
lations implied in the theorem are limitless.

This theorem with its many proofs is a strik-
ing illustration of the fact that there is more than
one vaj of establishing the same truth.

But before proceeding to the methods of dem-
onstration, the following historical account trans-
lated from a monograph by Jury Wlpper, published in
1880, and entitled ”46 Bewelse des Pythagoralschen
Lehrsatzes," may prove both interesting and profita-
ble.

5

4

THE PYTHAGOREAN PROPOSITION

Wlpper acknowledges his indebtedness to
P. Graap who translated it out of the Russian, It is
as follows: "One of the weightiest propositions in
geometry if not the weightiest with reference to its
deductions and applications is doubtless the so-
called Pythagorean proposition."

The Greek text is as follows:

'Ev Tot'c 6p6oYwvt^otc 6016 xrjc tt^v 6p0r^v
ywvi'av uTcoTEi vouar)c xXeupSc TexpaY^vov taov iarC xoTc
ind xSv xrfv 6p67^v Tceptex^^^wv xXeupSv

xexpaycSvo i c.

The Latin reads: In rectangulis triangulis
quadratum, quod a latere rectum angulum subtendente
describitur, aequale est els, quae a laterlbus rectum
angulum continentibus descrlbuntur ,

German: In den rechtwinkellgen Drelecken 1st

das Quadrat, welches von der dem rechten Wlnkel
gegenuber liegenden Seite beschrleben Wlrd, den Quad-
raten, welche von den ihn umschllessenden Seiten
beschrleben werden, glelch.

According to the testimony of Proklos the
demonstration of this proposition is due to Euclid
who adopted it in his elements (l, 47). The method
of the Pythagorean demonstration remains unknown to
us. It is undecided whether Pythagoras himself dis-
covered this characteristic of the right triangle, or
learned it from Egyptian priests, or took it from
Babylon: regarding this opinions vary.

According to that one most widely disseminat-
ed Pythagoras learned from the Egyptian priests the
characteristics of a triangle in which one leg = 3
(designating Osiris), the second = 4 (designating
Isis), and the hypotenuse = 5 (designating Horus) :
for which reason the triangle Itself is also named
the Egyptian or Pythagorean.*

*(Note. The Grand Lodge Bulletin, A.P. and A.M., of Iowa, Vol.
30 , No. 2, Feh. 1929> P* 42, has: In an old Egyptian manu-
Borlpt, recently discovered at Eedian, and supposed to belong

THE PYTHAGOREAN PROPOSITION

5

The characterletlcs of such a triangle, how-
ever, were laiown not to the Egyptian priests alone,
the Chinese scholars also knew them. "In Chinese
history," says Mr. Skatschkow, "great honors are
awarded to the brother of the ruler Uwan, Tschou-Gun,
who lived 1100 B.C.: he knew the characteristics of
the right triangle, (perfected) made a map of the
stars, discovered the compass and determined the
length of the meridian and the equator.

Another scholar (Cantor) says: this emperor
wrote or shared in the composition of a mathematical
treatise in which were discovered the fundamental
features, ground lines, base lines, of mathematics,
in the form of a dialogue between Tschou-Gun and
Schau-Gao. The title of the book is: Tschaou pi,
l.e., the high of Tschao. Here too are the sides of
a triangle already named legs as in the Greek, Latin,
German and Russian languages.

Here are some paragraphs of the 1st chapter
of the work. Tschou-Gun once said to Schau-Gao: "I
learned, sir, that you know numbers and their appli-
cations, for which reason I would like to ask how old
Po-chi determined the degrees of the celestial sphere.
There are no steps on which one can climb up to the
sky, the chain and the bulk of the earth are also in-
applicable; I would like for this reason, to knowhow
he determined the numbers."

Schau-Gao replied: "The art of counting goes
back to the circle and square."

If one divides a right triangle into its
parts the line which unites the ends of the sides

(Footnote continued) to the time of the Twelfth Pynasty, we
find the following equations: 1* + (f)® = (l^)®; 8® + 6®

- 10®; 2® + (1^)® = (2|)®; l6®+ 12® = 20®; all of which are
forms of the 3-^-5 triangle We also find that this tri-

angle was to them the synibol of universal nature. The base k
represented Osiris; the perpendicular 3# Isis; and the hypote-
nuse represented Horus, their son, being the product of the
two principles, male and female.)

6

THE PYTHAGOREAN PROPOSITION

when the base =5, the altitude = 4 is 5-

Tschou-Gvin cried out: "That Is Indeed ex-
cellent , "

It Is to be observed that the relations be-
tween China and Babylon more than probably led to the
assumption that this characteristic was already known
to the Chaldeans. As to the geometrical demonstra-
tion It comes doubtless from Pythagoras himself. In
busying with the addition of the series he could very
naturally go from the triangle with sides 5, 4 and 5,
as a single Instance to the general characteristics
of the right triangle.

After he observed that addition of the series
of odd number (1+3=4, 1+3+5= 9, etc.) gave
a series of squares, Pythagoras formulated the rule
for finding, logically, the sides of a right triangle:
Take an odd number (say 7) which forms the shorter
side, square It (7^ = 49), subtract one (49 - 1 = 48),
halve the remainder (48 - 2 = 24); this half Is the
longer side, and this Increased by one (24 + 1 = 25),
Is the hypotenuse.

The ancients recognized already the signifi-
cance of the Pythagorean proposition for which fact
may serve among others as proof the account of Dioge-
nes Laertius and Plutarch concerning Pythagoras, The
latter Is said to have offered (sacrificed) the Gods
an ox In gratitude after he learned the notable char-
acteristics of the right triangle. This story Is
without doubt a fiction, as sacrifice of animals,
l.e,, blood- shedding, antagonizes the Pythagorean
teaching.

During the middle ages this proposition which
was also named tnventum hecatombe diinum (in-as-much
as It was even believed that a sacrifice of a heca-
tomb — 100 oxen — was offered) won the honor -designa-
tion Maiiater matheaeoa, and the knowledge thereof
was some decades ago still the proof of a solid mathe-
matical training (or education). In examinations to
obtain the master’s degree this proposition was often
given; there was Indeed a time, as Is maintained.

THE PYTHAGOREAN PROPOSITION

7

when from every one who submitted himself to the test
as master of mathematics a new (original) demonstra-
tion' was required.

This latter circumstance, or rather the great
significance of the proposition under consideration
was the rea‘^on why numerous demonstrations of It were
thought out.

The collection of demonstrations which we
bring In what follows,* must. In our opinion, not
merely satisfy the simple thirst for knowledge, but
also as Important aids In the teaching of geometry.
The variety of demonstrations, even when some of them
are finical, must demand In the learners the develop-
ment of rigidly logical thinking, must show them how
many sldedly an object can be considered, and spur
them on to test their abilities In the discovery of
like demonstrations for the one or the other proposi-
tion. ”

Brief Biographical information
Concerning Pythagoras

"The birthplace of Pythagoras was the Island
of Samos; there the father of Pythagoras, Mnessarch,
obtained citizenship for services which he had ren-
dered the Inhabitants of Samos during a time of fam-
ine. Accompanied by his wife Plthay, Mnessarch fre-
quently traveled In business Interests; during the
year 569 A.C. he came to Tyre; here Pythagoras was
born. At eighteen Pythagoras, secretly, by night,
went from (left) Samos, which was In the power of the
tyrant Polycrates, to the Island Lesbos to his uncle
who welcomed him very hospitably. There for two years
he received Instruction from Perekld who with Anak-
slmander and Thales had the reputation of a philoso-
pher.

*Not0. There were hut kS different demonstrations in the mono-
graph hy Jury Wipper, which 46 are among the classified collec-
tion found in this work.

8

THE PYTHAGOREAN PROPOSITION

After Pythagoras had made the religious ideas
of his teacher his own, he went to Anaksimander and
Thales in Miletus (5^9 A.C.). The latter was then
already 90 years old. With these men Pythagoras stud-
ied chiefly cosmography, i.e.. Physics and Mathemat-
ics.

Of Thales it is known that he borrowed the
solar year from Egypt; he knew how to calculate sun
and moon eclipses, and determine the elevation of a
pyramid from its shadow; to him also are attributed
the discovery of geometrical projections of great im-
port; e.g., the characteristic of the angle which is
inscribed and rests with its sides on the diameter,
as well as the characteristics of the angle at the
base of an (equilateral) Isosceles triangle.

Of Anaksimander it is known that he knew the
use of the dial in the determination of the sun*s ele-
vation; he was the first who taught geography and
drew geographical maps on copper. It must be observed
too, that Anaksimander was the first prose writer, as
down to his day all le^ngd works were written in
verse, a procedure which continued longest among the
East Indians. ^

Thales directed the eager youth to Egypt as
the land where he could satisfy his thirst for knowl-
edge. The Phoenician priest college in Sldon must in
some degree serve as preparation for this journey.
Pythagoras spent an entire year there and arrived in
Egypt 5^7.

Although Polikrates who had forgiven Pytha-
goras* nocturnal flight addresses to Amasls a letter
in which he commended the young scholar, it cost
Pythagoras as a foreigner, as one unclean, the most
incredible toll to gain admission to the priest caste
which only unwillingly initiated even their own peo-
ple into their mysteries or knowledge.

The priests in the temple Heliopolis to whom
the king in person brought Pythagoras declared it im-
possible to receive him into their midst, and direct-
ed him to the oldest priest college at Memphis, this

PYTHAGORAS

Fro® a Freseo Eaplia«l.

THE POTHAGOREAN PROPOSITION

9

commended him to Thebes, Here somewhat severe condi-
tions were laid upon Pythagoras for his reception
into the priest caste; but nothing could deter him.
Pythagoras performed all the rites, and all tests,
and his study began under the guidance of the chief
priest Sonchis.

During his 21 years stay in Egypt Pythagoras
succeeded not only in fathoming and absorbing all the
Egyptian but also became sharer in the highest honors
of the priest caste.

In 527 Amasls died; in the following (526)
year in the reign of Psanimenlt, son of Amasls, the
Persian king Kambis invaded Egypt and loosed all his
fury against the priest caste.

Nearly all members thereof fell into captivi-
ty, among them Pythagoras, to whom as abode Babylon
was assigned. Here in the center of the world com-
merce where Bactrians, Indians, Chinese, Jews and
other folk came together, Pythagoras had during 12
years stay opportunity to acquire those learnings in
which the Chaldeans were so rich.

A singular accident secured Pythagoras liber-
ty in consequence of which he returned to his native
land in his 56th year. After a brief stay on the
Island Delos where he fo\ind his teacher Perekld still
alive, he spent a half year in a visit to Greece for
the purpose of making himself familiar with the re-
ligious, scientific and social condition thereof.

The opening of the teaching activity of Pytha
goras, on the Island of Samos, was extraordinarily
sad; in order not to remain wholly without pupils he
was forced even to pay his sole pupil, who was also
named Pythagoras, a son of Eratokles, This led him
to abandon his thankless land and seek a new home in
the highly cultivated cities of Magna Graecla (Italy)

In 510 Pythagoras came to Kroton. As is
known it was a turbulent year. Tarquln was forced to
flee from Rome, Hlpplas from Athens; in the neighbor-
hood of Kroton, in Slbarls, insurrection broke out.

The first appearance of Pythagoras before the
people of Kroton. began with an oration to the youth

10

THE PYTHAGOREAN PROPOSITION

wherein he rigorously but at the same time so con-
vincingly set forth the duties of young men that the
elders of the city entreated him not to leave them
without guidance (counsel). In his second oration
he called attention to law abiding and purity of mor-
als as the butresses of the family. In the two fol-
lowing orations he turned to the matrons and chil-
dren. The result of the last oration In which he
specially condemned liixury was that thousands of
costly garments were brought to the temple of Hera,
because no matron could make up her mind to appear
In them on the street.

Pythagoras spoke captlvatlngly, and It Is for
this reason not to be wondered at that his orations
brought about a change In the morals of Kroton*s In-
habitants; crowds of listeners streamed to him. Be-
sides the youth who listened all day long to his
teaching some 600 of the worthiest men of the city,
matrons and maidens, came together at his evening
entertainments; among them was the young, gifted and
beautiful Theana, who thought It happiness to become
the wife of the 60 year old teacher.

The listeners divided accordingly Into disci-
ples, who formed a school In the narrower sense of
the word, and Into auditors, a school In the broader
sense. The former, the so-called mathematicians were
given the rigorous teaching of Pythagoras as a scien-
tific whole In logical succession from the prime con-
cepts of mathematics up to the highest abstraction of
philosophy; at the same time they learned to regard
everything fragmentary In knowledge as more harmful
than Ignorance even.

Prom the mathematicians must be distinguished
the auditors (university extensioners) out of whom
subsequently were formed the Pythagoreans. These
took part In the evening lectures only In which noth-
ing rigorously scientific was taught. The chief
themes of these lectiires were: ethics. Immortality
of the soul, and transmigration — metempsychology.

About the year 490 when the Pythagorean
school reached Its highest splendor- -brilliancy — a

THE PYTHAGOREAN PROPOSITION

11

certain Hypasos who had been expelled from the school
as unworthy put himself at the head of the democratic
party In Kroton and appeared as accuser of his former
colleagues. The school was broken up, the property
of Pythagoras was confiscated and he himself exiled.

The subsequent l6 years Pythagoras lived In
Tarentum, but even here the democratic party gained
the upper hand In ^7^ and Pythagoras a 95-y©ar old
man must flee again to Metapontus where he dragged
out his poverty-stricken existence 4 years more. Fi-
nally democracy triumphed there also; the house In
which was the school was burned, many disciples died
a death of torture and Pythagoras himself with dif-
ficulty having escaped the flames died soon after In
his 99th year."*

Supplementary Historical Data

To the following (Graap’s) translation, out
of the Russian, relative to the great master Pytha-
goras, these Interesting statements are due,

"Fifteen hundred years before the time of
Pythagoras, (549-470 B.C.),** the Egyptians construct-
ed right angles by so placing three pegs that a rope
measured off Into 4 and 5 units would just reach
around them, and for this purpose professional ’rope
fasteners’ were employed.

"Today carpenters and masons make right an-
gles by measuring off 6 and 8 feet In such a manner
that a *ten-foot pole’ completes the triangle.

"Out of this simple Nlle-compelllng problem
of these early Egyptian rope-fasteners Pythagoras Is
said to have generalized and proved this Important
and famous theorem, — the square upon the hypotenuse

*Note. The above translation is that of Rr. Theodore H. John-
ston, Principal (190?) of the West High School, Cleveland, 0.

**Hfote. From recent accredited hlographlcal data as to Pytha-
goras, the record reads: ”Bom at Samos, c. 5^2 B.C. Died
probably at Metaponttim, c. 501, B.C.”

12

THE PYTHAGOREAN PROPOSITION

of a right triangle Is equal to the sum of the
squares upon Its tvo legs, — of which the right tri-
angle whose sides are 5, ^ and 5 Is a simple and par-
ticular case; and for having proved the universal
truth Implied In the triangle, he made his name

Immortal- -written Indelibly across the ages.

In speaking of him and his philosophy, the
Journal of the Royal Society of Canada, Section II,
Vol. 10, 1904, p. 239, says: "He was the Newton, the
Galileo, perhaps the Edison and Marconi of his

Epoch 'Scholars now go to Oxford, then to Egypt,

for fundamentals of the past The philosophy of

Pythagoras Is Aslatlc--the best of Indla--ln origin.

In which lore he became proficient; but he committed
none of his views to writing and forbid his followers
to do so. Insisting that they listen and hold their
tongues. ' "

He was Indeed the Sarvonarola of his epoch;
he excelled In philosophy, mysticism, geometry, a
writer upon music, and In the field of astronomy he
anticipated Copernicus by making the sun the center
of the cosmos. "His most original mathematical work
however, was probably In the Greek Arlthmetlca, or
theory of numbers, his teachings being followed by
all subsequent Greek writers on the subject."

Whether his proof of the famous theorem was
wholly original no one knows; but we now know that
geometers of Hindustan knew this theorem centuries
before his time; whether he knew what they knew Is
also unknown. But he, of all the masters of antiqui-
ty, carries the honor of Its place and Importance In
our Euclidian Geometry.

On account of Its extensive application In
the field of trigonometry, surveying, navigation and
astronomy. It Is one of the most. If not the most.
Interesting propositions In elementary plane geometry.

It has been variously denominated as, the
Pythagorean Theorem, The Hecatomb Proposition, The
Carpenter's Theorem, and the Pons Aslnorum because of
Its supposed difficulty. But the term "Pons Aslnorum"

THE PYTHAGOREAN PROPOSITION

13

also attaches to Theorem V, properly, and to Theorem
XX erroneously, of Book I of Euclid* s Elements of
Geometry.

It Is regarded as the most fascinating Theo-
rem of all Euclid, so much so, that thinkers from all
classes and nationalities, from the aged philosopher
In his armchair to the young soldier in the trenches
next to no-man* s -land, 1917, have vhlled away hours
seeking a new proof of Its truth.

Camerer, in his notes on the First Six Books
of Euclid* s Elements gives a collection of 17 differ-
ent demonstrations of this theorem, and from time to
time others have made collections, --one of 28, an-
other of 33, Wlpper of 46, Versluys of 96, the Ameri-
can Mathematical Monthly has 100, others of lists
ranging from a few to over 100, all of which proofs,
with credit, appears In this (now, 19^0) collection
of over 360 different proofs, reaching In time, from
900 B.C. , to 1940 A.D.

Some of these 367 proof s, --supposed to be
new- -are very old; some are short and simple; others
are long and complex; but each Is a way of proving
the same truth.

Read and take your choice; or better, find a
new, a different proof, for there are many more proofs
possible, whose figure will be different from any
one found herein.

*Note. Perhaps J.G. See Notes and Queries, 1879, Vol. V, No.
4l, p. 41.

"Mathematics is queen of the sci-
ences and arithmetic is queen of Mathe-
matics* She often condescends to render
service to astronomy and other natural
sciences^ but under all circumstances the
first place is her due."

Gauss ( 1777 - 1855 )

THE PYTHAGOREAN THEOREM
From an Ar i thmet i co-AI gebra i c Point of View

Dr. J. W. L. Glashler In his address before
Section A of the British Association for the Advance-
ment of Science, I 89 O, said: ”Many of the greatest
masters of the Mathematical Sciences were first at-
tracted to mathematical Inquiry by problems concern-
ing numbers, and no one can glance at the periodicals
of the present day which contains questions for solu-
tion without noticing how singular a charm such prob-
lems continue to exert.”

One of these charming problems was the deter-
mination of ”Triads of Arithmetical Integers” such
that the sum of the squares of the two lesser shall
equal the square of the greater number.

These triads, groups of three, represent the
three sides of a right triangle, and are Infinite In
number .

Many ancient master mathematicians sought
general formulas for finding such groups, among whom
worthy of mention were Pythagoras (c. 582-c. 501 B.C.),
Plato (429-348 B.C.), and Euclid (living 300 B.C.),
because of their rules for finding such triads.

In our public libraries may be found many
publications containing data relating to the s\am of
two square numbers whose sim Is a square number among
which the following two mathematical magazines are
especially worthy of notice, the first being ”The
Mathematical Magazine,” 1891^ Vol. II, No. 5# In
which, p. 69 , appears an article by that master Mathe-
matical Analyst, Dr. Artemas Martin, of Washington,
D.C.; the second being ”The American Mathematical
Monthly,” 1894, Vol. I, No. 1, In which, p. 6, ap-
pears an article by Leonard E. Dickson, B.Sc., then
Fellow In Pure Mathematics, University of Texas.

17

18

THE PYTHAGOREAN PROPOSITION

Those who are Interested and desire more data
relative to such numbers than here culled therefrom,
the same may he obtained from these two Journals.

Prom the article by Dr. Martin: "Any number
of square numbers whose sum Is a square number can be
found by various rigorous methods of solution."

Case I . Let It be required to find two
square niimbers whose Siam Is a square niomber.

First Method . Take the well-known Identity
(x + y)^ = + 2xy -f y2 = (x - y)^ + 4xy. (l)

Now If we can transform 4xy Into a square we
shall have expressions for two square numbers whose
sum Is a square number.

Assume x = mp^ and y = mq^, and we have
4xy = 4m^p^q^, which Is a square number for all val-
ues of m, p and q; and (l) becomes, by substitution,
(mp^ + mq^)^ = (mp^ - mq^)^ + (2mpq)^, or striking
out the common square factor m^, we have (p^ + q^)^

= (P® - q®)® + (2pq)®. —(2)

Dr. Martin follows this by a second and a
third method, and discovers that both (second and
third) methods reduce, by simplification, to formula
( 2 ).

Dr. Martin declares, (and supports his decla-
ration by the Investigation of Matthew Collins*

"Tract on the Possible and Impossible Cases of Quad-
ratic Duplicate Equalities In the Dlophantlne Analy-
sis," published at Dublin In I858), that no expres-
sion for two square numbers whose sum Is a square can
be found which are not deduclble from this, or re-
ducible ^ this formula, — that (2pq)^ + (p^ - q^)^ Is
always equal to (p^ + q^)^.

His numerical Illustrations are:

Example 1 . Let p = 2, and q = 1; then

p^ -f q® = 5, p2 - q2 3^ 2pq = 4, and we have 3^ + 4^

= 5^.

Example 2 . Let p = 3, q = 2; then p^ + q®

* 13, p2 - q2 = 5, 2pq = 12. 5^ + 12^ = 13®, etc.,

ad Inflnltijm.

THE PYTHAGOREAN THEOREM

19

Prom the article by Mr. Dickson: *Let the
three integers used to express the three sides of a
right triangle be prime to each other, and be symbol-
ized by a, b and h. * Then these facts follow:

1. They can not all be even numbers, otherwise they
would still be divisible by the common divisor 2.

2. They can not all be odd numbers. For a^ -f b^ = h^.
And if a and b are odd, their squares are odd, and
the sum of their squares is even; i.e., h^ is even.
But if h^ is even h must be even.

3. h must always be odd ; and, of the remaining two,
one must be even and the other odd. So two of the
three Integers, a, b and h, must always be odd,

(For proof, see p. 7, Vol. I, of said Am. Math.
Monthly. )

4 . When the sides of a right triangle are Integers,
the perimeter of the triangle is always an even
number, and its area is also an even number.

and h.

Rules for finding Integral values for a, b

Rule of Pythagoras : Let n be odd; then n.

For

n2 - 1

and ^ are three mich numbers.

n2 +

(=Mr-

4na + n* - 2n2 + 1

2. Platons Rule : Let m be any even number divisible
by 4 ; then m, ^ - 1, and ^ + 1 are three such

numbers. For m^ + - 1^^

El

16

mr

2

+ 1

3 . Euclid* s Rule : Let x and y be any two even or odd
numbers, such that x and y contain no common fac-
tor greater than 2, and xy is a square. Then >/Sc^
X — y X + y

— and t- ;;; are three such numbers. For

20

THE PYTHAGOREAN PROPOSITION

(v^)* + +

x® - 2xy +

4. Rule of Maseres (1721-1824): Let in and n be any

4- n^

two even or odd, m > n, and r an integer.

Thenni^, — and r are three such numbers.

2n 2n

„ ^ - n^ 4m^n^ + m'^ - 2m^ + n^ + n'^

For m2 + = —

2n 4n2

_ / m^ + n^ \^

“V 2n y *

5. Dickson* s Rule : Let m and n be any two prime in-
tegers, one even and the other odd , m > n and 2inn
a square. Then m + \/2mn, n 4- v/2mn and m 4- n
4 \/2mn are three such numbers. For (m 4- \/2nm)^

4 (n 4- /2mn) ^ 4 m^ 4 n^ 4 4mn 4 2m V^2mn 4 2n v^2mn
= (m 4 n 4 \/2mn)^.

6. By inspection it is evident that these five rules,
--the formulas of Pythagoras, Plato, Euclid,
Maseres and Dickson, — each reduces to the formula
of Dr. Martin.

In the Rule of Pythagoras: multiply by 4 and
square and there results (2n)^ 4 (n^ - 1)^ = (n^4 1)^,
in which p = n and q = 1.

In the Rule of Plato: multiply by 4 and
square and there results (2m) ^ 4 (m^ - 2^)^

= (m^ 4 2^)^, in which p = m and q = 2.

In the Rule of Euclid: multiply by 2 and
square there results (2xy)^ 4 (x - y)^ = (x 4 y)^, in
which p = X and q = y.

In the Rule of Maseres: multiply by 2n and
square and results are (2mn)^ 4 (m® - n^)^

* (m^ 4 n^)^, in which p = m and q * n.

In Rule of Dickson: equating and solving

and

Or If desired, the formulas of Martin, Pytha-
goras, Plato, Euclid and Maseres may be reduced to
that of Dickson.

The advantage of Dickson* s Rule Is this: It
gives every possible set of values for a, b and h In
their lowest terms, and gives this set but once.

To apply his rule, proceed as follows: Let
m be any odd square whatsoever, and n be the double
of any square number whatsoever not divisible by m.

Examples . If m = 9> n may be the double of
1, 4, l6, 2^ 49, etc.; thus when m = 9, a nd n = 2,
then m + v/2mn = 15, n + \/2mn =8, m 4- n + v/2mn = 17.
So a = 8, b * 15 and h = 17.

If m = 1, and n = 2, we get a = 3, b = 4,

h = 5.

If m = 25, and n = 8, we get a = 25, b = 45,
h = 53, etc. , etc.

Tables of Integers for values of a, b and h
have been calculated.

Halsted*s Table (in his "Mensuration") Is ab-
solutely complete as far as the 59th set of values.

METHODS OF PROOF

Method ts the following of one thing
throug^h another. On^eL the following of
one thing after, another.

The type and form of a figure necessarily de-
termine the possible argument of a derived proof;
hence, as an aid for reference, an order of arrange-
ment of the proofs Is of great Importance.

In this exposition of some proofs of the
Pythagorean theorem the aim has been to classify and
arrange them as to method of proof and type of fig-
ure used; to give the name. In case It has one, by
which the demonstration Is known; to give the name
and page of the journal, magazine or text wherein the
proof may be found. If known; and occasionally to
give other Interesting data relative to certain
proofs.

The order of arrangement herein Is, only In
part, my own, being formulated after a study of the
order found In the several groups of proofs examined,
but more especially of the order of arrangement given
In The American Mathematical Monthly, Vols. Ill and

IV, 1896-1899.

It Is assumed that the person using this work
will know the fundamentals of plane geometry, and
that, having the figure before him, he will readily
supply the "reasons why" for the steps taken as,
often from the figure, the proof Is obvious; there-
fore only such statements of construction and demon-
stration are set forth In the text as Is necessary to
establish the agrument of the particular proof.

22

ALGEBRAIC PROOFS

23

The Methods of Proof Are;

I. ALGEBRAIC PROOFS THROUGH LINEAR RELATIONS

A. Similar Riiht Trianiles

From linear relations of similar right trian-
gles it may he proven that. The square of the hypote-
nuse of a riiht trianile is equal to the sum of the
squares of the other two sides.

And since the algebraic square is the measure
of the geometric square, the truth of the proposition
as just stated involves the truth of the proposition
as stated under Geometric Proofs through comparison
of areas. Some algebraic proofs are the following;

One

In rt, trl. ABH, draw HC
perp. to AB. The tri’s ABH, ACH and
HCB are similar. For convenience,
denote BH, HA, AB, HC, CB and AC by
a, b, h, X, y and h-y resp*y. Since,
from three similar and related tri-
angles, there are possible nine sim-
ple proportions, these proportions
and their resulting equations are;

(1) a ; X = b : h - y /. ah - ay = bx.

(2) a ; y = b ; x /. ax = by.

(3) X : y = h - y ; x x® = hy - y®.

(4) a : X = h : b ab = hx.

(5) a ; y = h ; a a^ = hy.

(6) X : y = b : a /. ax = by.

(7) b : h - y = h : b b® = h® - hy.

(8) b : X = h ; a /. ab = hx.

(9) h-y : X « b : a /. ah - ay = bx. See Versluys,
p. 86, fig. 97, ¥m. ¥. Rupert,

Since equations (l) and (9) are identical,
also (2) and (6), and (4) and (8), there remain but
six different equations, and the problem becomes.

24

THE PYTHAGOREAN PROPOSITION

how may these six equations be combined so as to give
the desired relation h^ = a^ + b^, which geometrical-
ly Interprested Is AB^ = BH^ + HA^.

In this proof One, and In every case here-
after, as In proof Sixteen, p. 4l, the symbol AB^, or
a like symbol, signifies AB^.

Every rational solution of h^ = a® + b^ af-
fords a Pythagorean triangle. See "Mathematical Mon-
ograph, No. l6, Dlophantlne Analysis," (1915) ^ by
R. D. Carmichael.

Ist, — Leiendre 's Solution

a. Prom no single equation of the above nine
can the desired relation be determined, and there Is
but one combination of two equations which will give
it; viz., (5) a^ = hy; (7) b^ = h^ - hy; adding these
gives h^ = a^ + b^.

This is the shortest proof possible of the
Pythagorean Proposition.

b. Since equations (5) and (7) are implied in
the principle that homologous sides of similar tri-
angles are proportional it follows that the truth of
this Important proposition Is but a corollary to the
more general truth — the law of similarity .

c. See Davis Legendre, I 858 , p. 112,

Journal of Education, I 888 , V. XXV^, p. 4o4,

fig. V.

Heath's Math. Monograph, 1900, No. 1, p.

19^ proof III, or any late text on
geometry.

d. W. W. Rouse Ball, of Trinity College, Cam-
bridge, England seems to think Pythagoras knew of
this proof.

2nd. — Other Solutions

a. By the law of combinations there are pos-
sible 20 sets of three equations out of the six dif-
ferent equations. Rejecting all sets containing (5)
and (7), and all sets containing dependent equations,
there are remaining 13 sets from which the elimina-
tion of X and y may be accomplished In 44 different

ALOEBRAIC PROOFS

25

ways, each giving a distinct proof for the relation
h® = a® + h®.

b. See the American Math. Monthly, I896,

V. Ill, p. 66 or Edward's Geometry, p. 157> fig. 15.

Fig. 2

( 1 )

( 2 )

(5)

(M

(5)

( 6 )

Produce AH to C so that OB
will be perpendicular to AB at B.

Denote BH, HA, AB, BC and
CH by a, b, h, x and y resp'y.

The triangles ABH, CAB and
BCH are similar.

Prom the continued propor-
tion b:h:a = a:x:y = h:b
+ y : X, nine different simple pro-
portions are possible, viz.

a ; X.

(7)

a : X = h : b + y.

a ; y.

(8)

a : y = h : X.

X : y.

(9)

X : b + y = y : X, from

h : b + y.

which six different

h : X.

equations are possible

b + y : X.

as in One above.

Ist, — Solutions From Sets of Two Equations

a. As In One, there Is but one set of two
equations, which will give the relation h^ a^ + b^.

b. See Am. Math. Mo., V. Ill, p. 66.

2nd. — Solution From Sets of Three Equations

a. As In 2nd under proof One, fig. 1, there
are 13 sets of three eq*s, giving 44 distinct proofs
that give h^ = a^ + b^.

b. See Am. Math. Mo., V. Ill, p. 66.

c. Therefore from three similar rt. trl*s so
related that any two have one side in common there
are 90 ways of proving that h® = a® + b^.

26

THE PYTHAGOREAN PROPOSITION

Ihree

Fig. 3

( 1 )

( 2 )

(?)

Take BD = BH and at D draw
CD perp, to AB forming the two simi-
lar trl*s ABH and CAD.

a. Prom the continued propor-
tion a:x = b:h = h:b-x the
simple proportions and their result-
ing eq ' s are :
h - a ah - a^ = bx.

b - X ab - ax = hx.

a : X = b :
a : X = h :
b : h - a =

As there are but three equations and as each
equation contains the unknown x in the 1 st degree.

h : b - X /. b^ - bx = h^ - ah.

there
= a^

are

,2

possible but three solutions giving h

+ b

b. See Am. Math. Mo., V. Ill, p. 66 , and
Math. Mo., 1859, V. II, No. 2, Dem. Fig. 5, on p. 45
by Richardson.

d.xC

Fig. k

In Pig. 4 extend AB to
C making BC = BH, and draw CD
perp. to AC. Produce AH to D,
forming the two similar trl ’ s
ABH and ADC.

Prom the continued pro-
portion b ; h + a = a : X
= h : b + X three equations are
possible giving, as In fig. 5 ^
three proofs.

See Am. Math. Mo., V. Ill, p. 67 .

u

Fig. 5

Live

Dl’aw AC the bisector of the
angle HAB, and CD perp. to AB, form-
ing the similar trl*s ABH and BCD.
Then CB = a - x and DB = h - b.

ALGEBRAIC PROOFS

27

From the continued proportion h : a - x
=a :h-b=b :x three equations are possible civ-
Ing, as In fig. 3, three proofs for

a. Original vlth the author, Feb. 23, 1926.

Six

Through D, any pt. In either
leg of the rt. triangle ABE, draw
DC perp. to AB and extend It to E a
pt. In the other leg produced, thus
forming the four similar rt. trl’s
ABH, BEC, ACD and EHD. Prom
the continued proportion (AB = h)

: (be = a + x) : (ED = v)

; (DA = b - y) = (BH = a) :

(BC = h - z) : (DH = y) : (DC = w)

= (HA = b) : (CE = V + w) : (HE = x) ; (CA = z),

eighteen simple proportions and eighteen different
equations are possible.

Prom no single equation nor from any set of
two eq's can the relation h^ = a^ + b^ be found but
from combination of eq*s Involving three, four or
five of the unknown elements u, w, x, y, z, solutions
may be obtained.

1st. — Proofs From Sets Involvini Three Unknown Ele-
ments

a. It has been shown that there Is possible
but one combination of equations Involving but three
of the unknown elements, viz., x, y and z which will
give h^ = a^ + b^.

b. See Am. Math. Mo., V. Ill, p. 111.

2nd. — Proofs From Sets Involving Four Unknown Ele-
ments

a. There are possible 114 combinations In-
volving but four of the unknown elements each of
which will give h^ = a^ + b^.

b. See Am. Math. Mo., V. Ill, p. 111.

Fig. 6

28

THE POTHAGOREAN PROPOSITION

3rd, — Proofs From Seta Involving All Five Unknown
El ementa

a. Similarly, there are 47^9 combinations In-
volving all five of the \inknowns, from each of which
h® = a^ + can be obtained.

b. See Am. Math. Mo., V. Ill, p. 112.

c. Therefore the total no. of proofs from
the relations Involved In fig. 6 Is 4864.

Sfevea

\

N

\

> t

Fig. 7
7 Is 9728.

Produce AB to E, fig. 7,
and through E draw, perp. to AE,
the line CED meeting AH pro-
duced In C and HB produced in D,
forming the four similar rt.
tri»s ABH, DBE, CAE and CDH.

a. As In fig. 6, eigh-
teen different equations are
possible from which there are
also 4864 proofs.

b. Therefore the total
no. of ways of proving that h^

= a^ + b^ from 4 similar rt.
trl’s related as In flg*s 6 and

c. As the pt. E approaches the pt. B, fig. 7
approached fig. 2, above, and becomes fig. 2, when E
falls on B.

d. Suppose E falls on AB so that CE cuts HB
between H and B; then we will have 4 similar rt. tri’s
Involving 6 unknowns. How many proofs will result?

liflht

In fig, 8 produce BH to D, making BD = BA,
and E, the middle pt. of AD, draw EC parallel to AH,
and join BE, forming the 7 similar rt. triangles AHD,
ECD, BED, BEA, BCE, BHF and AEP, but six of which

ALGEBRAIC PROOFS

29

Fig. 8

need consideration, since trl*s BED
and BEA are congruent and. In sym-
bolization, Identical .

See Versluys, p. 87, fig. 98,
Hoffmann, I818.

Prom these 6 different rt.
triangles, sets of 2 trl*s may be
selected In 15 different ways, sets
of 5 trl*s may be selected In 20
different ways, sets of 4 trl*s may
be selected In 15 different ways, sets of 5 trl*s may
be selected In 6 different ways, and sets of 6 trl*s
may be selected In 1 way, giving. In all, 57 differ-
ent ways In which the 6 triangles may be combined.

But as all the proofs derivable from the sets
of 2, 3, 4, or 5 trl*s are also found among the
proofs from the set of 6 triangles, an investigation
of this set will suffice for all.

In the 6 similar rt. trl*s, let AB = h, BH
= a, HA = b, DE = EA = X, BE = y, PH = z and BP = v,
b h - a

whence EC =

DH

h - a, DC =

EP = y

BE =

h + a

AD = 2x and AP = b - z, and from these

data the continued proportion Is

b : b/2 ; y ; (h + a)/2 : a : x

= h - a : (h - a)/2 : x : b/2 : z : y - v

=2x:x:h:y:v:b-z.

Prom this continued proportion there result
45 simple proportions which give 28 different equa-
tions, and, as groundwork for determining the number
of proofs possible, they are here tabulated.

(1) b b/2 = h - a : (h - a)/2, where 1 = 1. Eq. 1.

(2) b : b/2 = 2x : X, whence 1=1. Eq. 1.

(3) h - a : (h - a)/2 = 2x : x, whence 1=1. Eq. 1®.

(4) b : y = h - a : X, whence bx = (h - a)y. Eq. 2.

(5) b : y = 2x : h, whence 2xy = bh. Eq. 3.

(6) h - a ; X = 2x : h, whence 2x® « h^ - ah. Eq. 4.

50

THE PYTHAGOREAN PROPOSITION

(7)

b :

(a + h)/2 =

h - a : b/2, whence b^ =

h^ -

a®.

Eq.

5.

(8)

b :

(h + a)/2 =

2x : y, whence (h •

f a)x :

= by.

Eq.

6.

(9)

h

a : b/2 = 2^?

: : y, whence bx =

(h - a

)y.

Eq.

2.

(10)

b :

a = h - a :

z, whence bz = (h

- a)a.

Eq.

7.

(11)

b :

a = 2x : V,

whence 2ax = bv. j

Eq. 8.

(12)

h -

a : z = 2x :

V, whence 2xz = (h - a)

V. Eq.

. 9.

(15)

b :

X = h - a :

y - V, whence (h -

a)x =

b(y -

V)

Eq.

10.

(1^)

b :

x=2x:b-

z, whence 2x^ = b

^ - bz

. Eq.

11.

(15)

h -

a : y - V =

2x : b - z, whence

2(y -

v)z

= (h - a) (b - z)

. Eq. 12.

(16)

b/2

: y = (h - a

.)/2 : X, whence bx

= (h

- a)y,

Eq.

2.

(17)

b/2

; y = X : h,

whence 2xy = bh.

Eq. 5

•

(18)

(h -

■ a)/2 ; X =

X : h, whence 2x^ =

= h= -

ah.

Eq.

4^.

(19)

b/2

: (h + a)/2

= (h - a)/2 : b/2.

whence b^

= h*

■ - a^. Eq.

5^.

(20)

b/2

: (h + a)/2

= X : y, whence (h

+ a)x

= by,

Eq.

6.

(21)

(h -

- a)/2 ; b/2

= X : y, whence bx

= (h

- a)y,

Eq.

2^.

(22)

b/2

: a = (h - a

,)/2 : z, whence bz

= (h

- a)a.

Eq.

7^.

(25)

b/2

: a = X : V,

whence 2ax = bv.

Eq. 8

2

(24)

(h ■

- a)/2 ; z =

X ; V, whence 2xz =

= (h -

a)v.

Eq.

9^.

(25)

b/2

: X = (h - a

.)/2 : y - V, whence (h -

a)x

= b(

[y - v). Eq.

10^.

(26)

b/2

: X = X ; b

- z, whence 2x^ = 1

- bz.

Eq.

11^.

(27)

(h -

- a)/2 : y -

V = X : b - z, whence 2(y - v)x

= (h - a) (b - z)

. Eq. 12^.

•

(28)

y :

(h + a)/2 =

X : b/2, whence (h

+ a)x

= by,

Eq.

6®.

(29)

y :

(h + a)2 = h

L : y, whence 2y^ =

h^ + 1

ah.

Eq.

15.

ALGEBRAIC PROOFS

31

(50) X : b/2 = h : y, whence 2xy = bh. Eq. 3®.

(31) y : a = X : z, whence ax = yz. Eq. l4.

(32) y : a = h : V, whence vy = ah. Eq. 15.

(33) X : z = h : V, whence vx = hz. Eq. I6.

(34) y : X = X : y - V, whence x® = y(y - v) . Eq. 17.

(35) y : X = h : b - z, whence hx = y(b - z). Eq. I8.

(36) x:y-v=h:b-z, whence (b - z)x
= h{y - v). Eq. I9.

(37) (h + a)/2 : a = b/2 : z, whence (h + a)z = ab.
Eq . 20 .

(38) (h + a)/2 : X = y : V, whence 2ay = (h + a)v.

Eq. 21.

(39) b/2 : z = y : V, whence 2yz = bv. Eq. 22.

(40) (h + a)/2 : x = b/2 : y - v, whence bx = (h + a)

(y - v). Eq. 23.

(41) (h + a)/2 : X = y : b - z, whence 2xy = (h + a)
(b - z). Eq. 24.

(42) b/2 :y-v=y:b-z, whence 2y(y - v) = b
= b^ - bz. Eq. 25.

(43) a ; X = z : y - V, whence xz = a(y - v). Eq. 26.

(44) a : X = V : b - z, whence vx = a(b - z). Eq. 27.

(45) z:y-v=v:b-z, whence v(y - v)

= (b - z)z. Eq. 28.

The symbol 2^, see (21), means that equation
2 may be derived from 4 dliTerent proportions. Simi-
larly for 6®, etc.

Since a definite no. of sets of dependent equa-
tions, three equations In each set. Is derivable from
a given continued proportion and since these sets
must be known and dealt with In establishing the no.
of possible proofs for h® = a^ + b^. It becomes nec-
essary to determine the no. of such sets. In any

continued proportion the symbolization for the no. of

such sets, three equations In each set. Is

In which n signifies the no. of simple ratios In a

member of the continued prop*n. Hence for the above

continued proportion there are derivable 75 such sets

of dependent equations. They are:

THE PYTHAGOREAN PROPOSITION

?2

(1),

(2), (5); (4), (5)

, (6)

; (7),

(8), (9); (10),

(11),

(12);

(15),

(14),

(15)

; (16)

, (17), (18);

(19),

(20),

(21);

(22),

(25),

(24)

; (25)

, (26), (27);

(28),

(29),

(50);

(51),

(52),

(55)

; (54)

, (55), (56);

(57),

(58),

(59);

(40),

(41),

(42)

; (45)

, (44), (45);

(1),

(4),

(16);

(1), (7), (19); (1), (10), (22); (1),

(15),

(25);

(4),

(7), (28); (4), (10), (51); (4), (15)

9

(?4);

(7),

(10),

(57);

(7),

(15),

(40); (10), (15),

(45);

(16),

(19),

(20);

(16)

, (22)

, (51); (16),

(25),

(54);

(19),

(22),

(57);

(19)

, (25)

, (40); (22),

(25),

(45);

(28),

(51),

(57);

(28)

, (54)

, (40); (51),

(54),

(45);

(57),

(40),

(45);

(2),

(5),

(17); (2), (8)

(20);

(2),

(11),

(25);

(2),

(14),

(26); (5), (8)

, (29);

(5),

(11),

(52);

VJl

IH

>

(55);

(8), (11), (58); (8),

(14),

(41);

(11),

(14),

(44)

; (17)

, (20), (29);

(17),

(25),

(52);

(17),

(26),

(55)

; (20)

, (25), (58);

(20),

(26),

(41);

(25),

(26),

(44)

; (29)

, (52), (58);

(29),

(55),

(41);

(52),

(55),

(44)

; (58)

, (41), (44);

(5),

(6),

(18);

(5), (9), (21); (5); (12), (24); (5),

(15),

(27);

(6),

(9), (50); (6), (12), (55); (6), (15)

9

(56);

(9),

(12),

(56);

(9),

(15),

(42); (12), (15),

(45);

(18),

(21),

(50);

(18)

, (24)

, (55); (18),

(27),

(56);

(21),

(24),

(59);

(21)

, (27)

, (42); (24),

(27),

(45);

(50),

(55),

(59);

(50)

, (56)

, (42); (55),

(56),

(45);

(59),

(42),

(45).

These 75 sets expressed In the symbolization
of the 28 equations give but 49 sets as follows:

1,

1, 1; 2,

3,

4;

2,

5, 6

; 7,

8,

9; 10

, 11

, 12; 6

,

15,

5; 14,

15,

16;

17

, 18

, 19

; 20

, 21,

22;

25

, 24

,

25;

26, 27,

28

; 1,

2,

2;

1, 5

, 5;

1, 7

, 7;

1,

10,

10

1,

6, 6; 2,

7,

14;

2,

10,

17;

5,

7, 20

; 5,

10

, 25

; 7

10,

26; 6,

14,

20;

6,

17,

25;

14,

17,

26;

20,

25,

26

1,

5, 5; 1,

8,

8;

1,

11,

11;

5, 8

, 15;

5,

11,

18;

6,

8,

21; 6, 11,

24;

8,

11,

27;

15,

15, 21; 15,

18,

24;

15,

18, 27;

21

, 24

, 27; 1

, 4,

4;

1, 9,

9;

1,

12,

12;

4,

9, 16; 4

, 12, 19;

2, 9

, 22

; 2,

12,

25;

9,

12,

28;

5,

16, 22;

3,

19,

25;

16,

19,

28;

22,

25,

28.

Since eq. 1 Is an Identity and eq. 5 gives,
at once, h^ = a^ + there are remaining 26 equa-
tions Involving the 4 unknowns x, y, z and v, and

ALGEBRAIC PROOFS

53

proofs may be possible from sets of equations Involv-
ing X and y, x and z, x and v, y and z, y and v, z
and V, X, y and z, x, y and v, x, z and v, y, z and v,
and X, y, z and v.

1st. — Proofs From Sets Involving Two Unknowns

a. The two unknowns, x and y, occur In the
following five equations, viz., 2, 5, 4, 6 and 15#
from which but one set of two, viz., 2 and 6, will
give h^ + a^ = h^, and as eq. 2 may be derived from
4 different proportions and equation 6 from 5 differ-
ent proportions, the no. of proofs from this set are
12 .

Arranged In sets of three we get,

2^, 5^# 13 giving 12 other proofs;

(2, 3# 4) a dependent set — no proof;

2^, 4^, 13 giving 8 other proofs;

(3# 6, 13) a dependent set — no proof;

3®, 4^, 6® giving 18 other proofs;

4^, 6®, 13 giving 6 other proofs;

3®, 4^, 13 giving 6 other proofs.

Therefore there are 62 proofs from sets In-
volving X and y.

b. Similarly, from sets Involving x and z
there are 8 proofs, the equations for which are 4, 7#
11, and 20.

c. Sets Involving x and v give no additional

proofs .

d. Sets Involving y and z give 2 proofs, but
the equations were used In a and b, hence cannot be
counted again, they are 7# 13 and 20.

e. Sets Involving y and v give no proofs.

f. Sets Involving z and v give same results

as d.

Therefore the no. of proofs from sets Involv-
ing two unknowns Is 70, making. In all 72 proofs so
far, since h® * a^ + b^ Is obtained directly from two
different prop’s.

34

THE PYTHAGOREAN PROPOSITION

2nd. — Proofs From Sets Involving Three Unknowns

a. The three unknowns x, y and z occur In
the following 11 equations, viz,, 2, 5, 4, 6, 7>
13, 14, 18, 20 and 24, and from these 11 equations

sets of four can be selected In

11 . 10 . 9 . 8

n

= 550

ways, each of which will give one or more proofs for
h® = a^ + b^. But as the 330 sets, of four equa-
tions each. Include certain sub-sets heretofore used,
certain dependent sets of three equations each found
among those In the above 75 sets, and certain sets of
four dependent equations, all these must be deter-
mined and rejected; the proofs from the remaining
sets will be proofs additional to the 72 already de-
termined.

Now, of 11 consecutive things arranged In

sets of 4 each, any one will occur In

9 . 8

120 of the 330 sets, any two In “

or

8

H"

10 . 9 . 8

Li

or 36 of the

330, and any three In j ^ or 8 of the 330 sets. There-

fore any sub-set of two equations will be found In
36, and any of three equations In 8, of the 330 sets.

But some one or more of the 8 may be some one
or more of the 36 sets; hence a sub- set of two and a
sub-set of three will not necessarily cause a rejec-
tion of 36 + 8 = 44 of the 330 sets.

The sub-sets which gave the 70 proofs are:

2, 6, for which 36 sets must be rejected;

7, 20, for which 35 sets must be rejected, since
7, 20, Is found In one of the 36 sets above;

2, 3, 13, for which 7 other sets must be rejected,
since

2, 3, 13, Is found In one of the 36 sets above;

2, 4, 13, for which 6 other sets must be rejected;

3, 4, 6, for which 7 other sets must be rejected;

4, 6, 13, for which 6 other sets must be rejected;

3, 4, 13, for which 6 other sets must be rejected;

4, 7, 11, for which 7 other sets must be rejected;

and

ALGEBRAIC PROOFS

55

4, 11, 20, for which 7 other sets must be rejected;

for all of which 117 sets must be rejected.

Similarly the dependent sets of three, which
are 2, 3, 4; 3, 6, 13; 2, 7, l4; 6, l4, 20; 3, 11,
l8; 6, 11, 24; and 13, l8, 24; cause a rejection of
6+6+6+6+8+7+8, or 47 more sets.

Also the dependent sets of four, and not al-
ready rejected, which are, 2, 4, 11, l8; 3, 7, 1^;

5, 6, 18, 24; 3, 13, 14, 20; 3, 11, 13, 24; 6, 11,

13, 18; and 11, l4, 20, 24, cause a rejection of 7
more sets. The dependent sets of fours are discovered
as follows: take any two dependent sets of threes
having a common term as 2, 3, and 3, 11, 18; drop
the common term 3, and write the set 2, 4, 11, l8; a
little study will disclose the 7 sets named, as well
as other sets already rejected; e.g., 2, 4, 6, 13.
Rejecting the 117 + 49 + 7 = 171 sets there remain
159 sets, each of which will give one or more proofs,
determined as follows. Write down the 350 sets, a
thing easily done, strike out the 171 sets which must
be rejected, and, taking the remaining sets one by
one, determine how many proofs each will give; e.g.,
take the set 2, 3, 7, H; write It thus 2*, 3®, 7®,
11®, the exponents denoting the different proportions
from which the respective equations may be derived;
the product of the exponents, 4x3x2x2= 48, is
the niimber of proofs possible for that set. The set
6®, 11^, 18^, 20^ gives 6 proofs, the set l4^, l8^,
20^, 24^ gives but 1 proof; etc.

The 159 sets, by investigation, give 1231

proofs.

b. The three unknowns x, y and v occur in the
following twelve equations, — 2, 3, 4, 6, 8, 10, 11,

13, 15, 17, 21 and 23, which give 495 different sets
of 4 equations each, many of which must be rejected
for same reasons as in a. Having established a method
in a, we leave details to the one interested.

c. Similarly for proofs from the eight equa-
tions containing x, z and v, and the seven eq*s con-
taining y, z and v.

36

THE PYTHAGOREAN PROPOSITION

3rd. — Proofs From Sets Involvini the Pour Unknowns
X, y, z and v.

a. The four unknowns occur In 26 equations;

hence there are « 6578O different

sets of 5 equations each. Rejecting all sets con-
taining sets heretofore used and also all remaining
sets of five dependent equations of which 2, 9#

19, 28, Is a type, the remaining sets will give us
many additional proofs, the determination of which
Involves a vast amount of time and labor If the meth-
od given In the preceding pages Is followed. If there
be a shorter method, I am unable, as yet, to discover
It; neither am I able to find anything by any other
Investigator.

Ath. — Special Solutions

a. By an Inspection of the 45 simple propor-
tions given above. It Is found that certain propor-
tions are worthy of special consideration as they
give equations from which very simple solutions fol-
low.

Prom proportions (7 ) and (19 ) h^ = a^ + b^
follows Immediately. Also from the pairs (4) and
(18), and (10) and (37), solutions are readily ob-
tained.

b. Hof fmann*3 solution.

Joh. Jos. Ign. Hof fmann made a collection of
32 proofs, publishing the same In "Der Pythagoralsche
Lehrsatz,” 2nd edition Mainz, 1821, of which the so-
lution from (7) l3 one. He selects the two triangles,
(see fig. 8), AHD and BCE, from which b : (h + a)/2
* h - a : b/2 follows, giving at once h® = a^ + b®.

See Jury Wlpper*s 46 proofs, I88O, p. 40, fig.
4l. Also see Versluys, p. 87, fig. 98, credited to
Hoffmann, 1818. Also see Math. Mo., Vol. II, No. II,
p. 45, as given In Notes and Queries, Vol. 5, No. 43,
p. 41.

c. Similarly from the two triangles BCE and
BCD b/2 : (h + a)/2 - (h - a)/2 : b/2, h® * a* + b^.

ALGEBRAIC PROOFS

37

Also from the three triangles AHD, BEA and BCE pro-
portions (4) and (8) follow, and from the three tri-
angles AHD, BHE and BCE proportions (lO) and (37)
give at once h^ = a^ +

See Am. Math. Mo., V. Ill, pp. 169-70.

Nine

Fig. 9

Produce AB to any pt.
D. Prom D draw DE perp. to
AH produced, and from E drop
the perp. EC, thus forming
the 4 similar rt. trl*s ABH,
AED , ECB and ACE .

Prom the homologous
sides of these similar tri-
angles the following con-
tinued proportion results:

(AH = b) : (AE = b + v) : (EC « w) : (AC = h + x)

= (BH = a) : (DE = y) : (CD = z) : (EC = w)

= (AB = h) : (AD = h + X + z) : (DE « y) : (AE = b + v).
Note--B and C do not coincide.

a. Prom this continued prop*n l8 simple pro-
portions are possible, giving, as In fig. 6, several
thousand proofs.

b. See Am. Math. Mo., V. Ill, p. 171.

Ista

In fig. 10 are three simi-
lar rt. trl*s, ABH, EAC and DEP,
from which the continued propor-
tion

(HA = b) ; (AC = h + v)

: (DP = DC = x)

= (HB = a) : (CE = y)'

; (PE = z) = (AB = h)

: (AE ■ h + V + z) ; (DE = y - x)

Fig. 10

38

THE PYTHAGOREAN PROPOSITION

follows giving 9 simple proportions from which many
more proofs for h^ = a® + may be obtained.

a. See Am. Math. Mo., V. Ill, p. 171.

Eleven

Fig. 11

From D in HH, so that DH
= DC, draw DC par. to HB and DE perp.
to AB, forming the 4 similar rt.
tri*s ABH, ACD, CDE and DAE, from
which the continued proportion

(BH = a) : (CD = DH = v ) : (EC = y)

: (DE = x) = (HA = b) : (DA = b - v)

: (DE = X ) : (AE = z ) = (AB = h )

: (AC = z + y) : (CD=v) : (AD = b - v)

follows; 18 simple proportions are possible from
which many more proofs for h^ = a^ + b^ result.

By an Inspection of the l8 proportions it is
evident that they give no simple equations from which
easy solutions follow, as was found in the investiga-
tion of fig. 8, as in a under proof Siiht.

a. See Am. Math. Mo., V. Ill, p. 171.

IwiLve

Fig. 12

The construction of fig. 12
gives five similar rt. triangles,
which are: ABH, AHD, HBD, ACB and
BCH, from which the continued
prop * n

(BH = a) : (HD = x) ; (BD = y)

: (CB = : (CH = •^) = (HA = b)

; (DA - h - y) ; (DH = x) ; (BA = h) : (HB = a)
= (AB = h) ! (AH = b) : (HB = a) : (AC = b +

ALCEBRAIC PROOFS

39

follows, giving 50 simple proportions from which only
12 different equations result. Prom these 12 equa-
tions several proofs for h^ = a^ + are obtainable.

a. In fig. 9f when C falls on B It Is obvious
that the graph become that of fig. 12. Therefore
the solution of fig. 12, Is only a particular case of
fig. 9; also note that several of the proofs of case
12 are Identical with those of case 1, proof One.

b. The above Is an original method of proof
by the author of this work.

ItLlrte^n

Fig. 13
DB : (BH =
and (2) HD

Complete the paral. and draw
HP perp. to, and EP par. with AB
resp*ly, forming the 6 similar trl*s,
BHA, HCA, BCH, AEB, DCB and DPE, from
which 45 simple proportions are ob-
tainable, resulting In several thou-
sand more possible proof for h^ = a^
+ b^, only one of which we mention.

(l) From tri*s DBH and BHA,

a^

a) = (BH = a) : (HA = b); /.DB = —

: (AB = h) = (BH = a) : (HA = b);

(5) Prom trl*s DPE and BHA,

DP : (EB - DB) - (BH = a) : (AB = h),
or DP : b^ - : a : h; DP = a^^ — .

(4) Trl.

, ABH = I

par. HE

= i AB X

HC = I ab

-

- i[»r ;

-

. b® - <
bh

^J)]

- ^ +

ab a®

A-h -

+ ab® - a'

“ TtT +

4 ■ 4b

• ^

ao = —

4b

40

THE PYTHAGOREAN PROPOSITION

whence (6) h^ = + h®.

a. This particular proof was produced by
Prof. D. A. Lehman, Prof, of Math, at Baldwin Uni-
versity, Berea, 0., Dec. 1899-

b. Also see Am. Math. Mo., V. VII, No. 10,

p. 228.

EQy.Lteen

Take AC and AD = AH
and draw HC and HD.

Proof. Trl’s CAH and
HAD are Isosceles. Angle CHD
Is a rt. angle, since A Is
equidistant from C, D and H.

Angle HDB = angle CHD
+ angle DCH.

= angle AHD + 2 angle CHA = angle CHB.

trl*s HDB and CHB are similar, having an-
gle DBH In common and angle DHB - angle ACH.

CB : EH = BH : DB, orh4-b:a = a:h-b.

Whence h^ = a® + b^.

a. See Math. Teacher, Dec., 1925. Credited
to Alvin Knoer, a Milwaukee High School pupil; also
Versluys, p. 85, fig. 95; also Encyclopadle der Ele-
mentar Mathematlk, von H. Weber \md J. Wellsteln,

Vol. II, p. 242, where, (1905), It Is credited to
C. G. Sterkenburg.

F i f teen

In fig. 15 the const *s Is
obvious giving four similar right
triangles ABH, AHE, HBE and HCD,
from which the continued proportion
(BH = a) : (HE = x) : (BE = y)

(CD - y/2) = (HA = b) : (EA = h - y)
(EH = x) : (DH = x/2) = (AB = h)
(AH = b) : (HB = a) : (HC = a/2)
follows, giving 18 slxgple proportions.

rig. 15

ALGEBRAIC PROOFS

41

a. Prom the two simple proportions

(1) a : y = h : a and

(2) b:h-y = h:bwe get easily = a^ + b^.

b. This solution is original with the author,
but, like cases 11 and 12, it is subordinate to case

1.

c. As the number of ways in which three or
more similar right triangles may be constructed so
as to contain related linear relations with but few
unknowns Involved is unlimited, so the number of pos-
sible proofs therefrom must be unlimited.

Sixteen

The two following proofs,
differing so much, in method, from
those preceding, are certainly
worthy of a place among selected
proofs.

Ist . — This proof rests on the
axiom, ”The whole is equal to the

It

Let AB = h, BH = a and HA = b, in the rt. trL
ABH, and let HC, C being the pt. where the perp. from
H Intersects the line AB, be perp. to AB. Suppose
h® = + b®. If h® = a® + b®, then a® = x® + y®

and b® = X® + (h - y)®, or h® = x® + y® + x® + (h -y
= y® + 2x® + (h - y)® = y® + 2y(h - y) + (h - y)®

= y + [(h - y)]®

h = y + (h - y), i.e., AB = BC + CA, which

is true.

the supposition is true, or h^ = a^ + b^.
a. This proof is one of Joh. Hof fmann^s 32
proofs. See Jury Wlpper, l880, p. 38, fig. 37

2 nd . — This proof is the "Reductlo ad Absurdum”

proof.

h^ <, *, or > (a^ + b^). Suppose it is less.

Fig. 16

mim nf* •nnT»+‘.a

42

THE PYTHAGOREAN PROPOSITION

Then, since h® = [(h - y) + y]® + [(h - y) + x®

+ (h - y)]® and = [ ax -f (h - y)]^, then

[(h - y) + X® + (h - y]® < [ax + (h - y)]® + a®.

[x= + (h - < a^[x2 + (h - y)^].

A a^ > X® + (h - y)*, which is absurd. For,
if the supposition be true, we must have a^ < x®

+ (h - y)^, as is easily shown.

Similarly, the supposition that h® > a® + b^,
will be proven false.

Therefore it follows that h® = a^ + b®.

a. See Am. Math. Mo., V. Ill, p, 170.

Seventeen

Take AE = 1, and draw EP

Fig, 17

perp, to AH, and HC perp. to AB.

HC = (AC X PE)/PE, BC = (HC x EE )/AP
= (AC X PE)/AP X FE/AP = AC X PEVaP®,
and AB = AC x CB = AC + AC x FE®/AP®

= AC(l + PE®)/AP® = AC(AP® + FE)®/AP®,

( 1 ).

But AB : AH = 1 : AP, whence AB = AH/AP, and
AH = AC/AP, Hence AB = AC/AP®. (2).

AC(AF® + EP®)/AP® = AC/AP®, AP® + PE® = 1.

.-. AB : 1 = AH ; AP. .-. AH = AB x AP. (3).

and BH = AB X PE. (4)

(3)® + (4)® = (5)®, or, AH® + BH® = AB® x AP® + AB®

X PE® = AB®(AP® + PE®) = AB®. .-. AB® = HB® + HA®, or
n = a + D .

a. See Math. Mo., (l859), Vol. II, No. 2,

Dem. 23, fig. 3-

b. An indirect proof follows. It is;

If AB® / (HB® + HA®), let x® = HB® + HA® then
X = (HB® + HA®)^^® = HA(l + HB®/HA®)*-'® - HA
(1 + PE®/PA®)^'® = HA [(PA® + PE®)/PA®]^^® = HA/PA
= AB, since AB ; AH = 1 : AP.

.-, if X = AB, X® = AB® = HB® + HA®. Q.E.D.

c. See said Math. Mo., (1859), Vol. II, No. 2,
Dem. 24, fig. 3.

ALGEBRAIC PROOFS

45

=

a® +

Prom slm. trl*s ABC and BCH,
HC = Angle ABC = angle CDA

= rt. angle. Prom sim. tri's AHD
and DHC, CD = ah/b; CB = CD. Area
of tri. ABC on base AC = ^ (b 4- a^/b)a.
Area of ACD on base AD = ^(ah/b)h.

(b + aVb)a = ahVb = (b^ + a^)/b

X a

ab + a''

a. See Versluys, p. 72, fig. 79*

Minel®®!!

Tri’s 1,

2 and 5 are sim-
ilar. Prom tri's

1 and 2, AC
= hVa, and CD
= hb/ a . Prom
tri* s 1 and 5,

EP = ha/b, and
PB = hVb.

Tri. CPH
= tri. 1 4- tri.

2 4- tri. 3 4“ sq.
AE.

So J (a 4“ h^/b ) (b 4- h^/a ) = Jab 4- Jh^ (b/ a ) 4- Jh^ (a/b )

4" h2, or a^b^ 4- 2abh^ 4* h'* = a^b^ 4* h^a 4- h^b + 2abh^,
or = h^a^ 4- h^b^. = a^ 4- b^. Q.E.D.

a. See Versluys, p. 23, fig. 80.

Iwentji

Draw HC perp. to AB and = AB. Join CB and
CA. Draw CD and CE nerp. resp’y to HB and HA.

44

THE PYTHAGOREAN PROPOSITION

Fig. 20

Area BHAC = area ABH
+ area ABC = But area trl.

CBH = ia®, and of trl. CHA =

= ^a® + l-b®. h® = a® + b®.
a. See Versluys, p. 75>
fig. 82, where credited to P. Armand
Meyer, I876.

Iwenty.-Qne

HC = HB = DE; HD = HA. Join
EA and EC. Draw EP and HG perp. to
AB and EK perp. to DC.

Area of trap. ABCD = area
(ABH + HBC + CHD + AHD ) = ab + Ja^

+ ib®. (1)

= area (EDA + EBC + ABE + CDE)

= Jab + Jab + (jAB x EP = jAB x AG

as trl '3 BEP and HAG are congruent)
= ab + J(AB = CD)(AG + GB) = ab + Jh®. (2)

ab + Jh® = ab + Ja® + Jb®. h® = a® + b®. Q.E.D.
a. See Versluys, p. 74, fig. 8l.

Iwent^-Iwo

In fig. 22, it Is obvious

that ;

(1) Trl. ECD = Jh®, (2) Trl. DBE
= (?) Trl. HAC =

/. (1) = (2) + (3) = (4) = ia^

+ = a^ + b^. Q.E.D.

a. See Versluys, p. 76, fig.
83, credited to Meyer, (1876); also
this work, p. 181, fig. 238 for a
similar geometric proof.

ALGEBRAIC PROOFS

45

Iwent^-Ih ree

For figure, use fig. 22 above, omitting lines
EC and ED. Area of sq. AD = (2 area of trl. DBH
= rect. BF) + (2 area of trl. HAC = rect. AF)

= 2 X + 2 X = a^ + b^ = h^. /. = a^ + b^.

Or use similar parts of fig. 315> In geometric proofs.

a. See Versluys, p. 76, proof 66, credited
to Meyer* 3, I876, collection.

Iwen ty.-Fgur

In fig. 22, denote HE by x. Area of trl. ABH
+ area of sq. AD = ^-hx + = area of (trl. ACH + trl.

CDH + trl. DBH) = Jb^ + |h(h + x) + ^a^ = 4b^ +

+ Jhx + ia^. = a^ + b^.

a. See Versluys, p. 76 , proof 67, and there
credited to P. Armand Meyer *s collection made In

1876.

b. Proofs Twenty-Two, Twenty-Three and Twenty-
Four are only variations of the Mean Proportional
Principle, --see p. 51, this book.

IweQ.tiL~Eive

Fig. 23

At A erect AC = to, and
perp. to AB; and from C drop (CD
= AH) perp. to AH. Join CH, CB and
DB. Then AD = HB = a. Trl. CDB
= trl. CDH = iCD X DH.

Trl. CAB =trl. CAD + trl.
DAB 4- (trl. BDC = trl. CDH = trl.
CAH + trl. DAB). Jh^ = ^a^ + ^b^.
. . h = a 4- b .

a. See Versluys, p. 77, fig. 84, one of
Meyer* 3, I876, collection.

Iwenty-Six

From A draw AC perp. to, and = to AB. Join
CB, and draw BF parallel and = to HA, and CD parallel
to AH and = to HB. Join CP and BD.

46

THE PYTHAGOREAN PROPOSITION

Trl. CBA = trl. BAP + trl.
FAC + trl. CBP = trl. BAP + trl. FAC
+ trl. PDB (alnce trl. EOF = trl.
EDB) = trl. FAC + trl. ADB. A ih®

= A + b^

a. See Versluys, p. 77 > fig.
85, being one of Meyer's collection.

Fig. 24

Iwen t^-Sgyen

Prom A draw AC perp. to,
and = to AB. From C draw CP equal
to HB and parallel to AH. Join CB,
AP and HP and draw BE parallel to
HA. CP = EB = BH = a. ACF and ABH
are congruent; so are CPD and BED.

Quad. BHAC = trl. BAC + trl.
ABH = trl. EBH + trl. HPA + trl. ACF
+ trl. FCD + trl. DBE. + Jab

= + Jb^ + l-ab . /. = a^ + b^ .

Fig. 25

86 ; also see
Vaes .

Q.E.D.

a. See Versluys^ p. 78, fig.
"Vrlend de Wlskunde," I898, by F. J.

Draw PHK perp. to AB
and make PH = AB. Join PA,

PB, AD and GB.

Trl * s BDA and BHP are
congruent; so are trl's GAB
and AHP . Quad . AHBP = trl .

BHP + trl. AHP. A ih® = Ja®

+ ib®. A h® = a® + b®. Q.E.D.

a. See Versluys, p. 79 ,
fig. 88 . Also the Sclentlflque
Revue, Feb. I6, I889, H. Renan;

ALGEBRAIC PROOFS

_47

1879-

also Pourrey, p. 77 and p. 99, — jal de Vulbert,

80 .

Iw£nt)^-Nj[ne

Through H draw PK perp. to
AB, making PH = AB, and join PA and
PB.

Since area AHBP = [area PHA
+ area PHB = ^h x aK + Jh x bK
= ih(AK + BK) = ih X h = ^h^] = (area
AHP + area BHP = + l-a^). /. ^h^

= "pSi 4- pO . . . h = a + b .

a. See Versluys, p. 79, fig.
89, being one of Meyer* s, I876, col-
lection.

Ihirti

Draw PH perp. to AB, making
PH = CD = AB. Join PA, PB, CA and
CB.

Tri. ABC = (tri. ABH + quad,
AHBC) = (quad. AHBC + quad. ACBP),
since PC = HD. In tri, BHP, angle
BHP = 180° - (angle BHD = 90° + angle
HBD). So the alt, of tri. BHP from

and its

,,2

area

lo2.

the vertex P = a,
likewise tri. AHP = But as in

fig. 27 above, area AHBP = ^h^. h

= a^ + b^. Q.E.D.

a. See Versluys, p.
Meyer* s, I876, collections.

80, fig. 90, as one of

Ih l.rt^-0ne

Tri *3 ABH and BDH are similar, so DH = a^/h
and DB = ab/h. Tri. ACD = 2 tri. ABH + 2 tri. DBH.

48

THE PYTHAGOREAN PROPOSITION

Area of tri. ACD = ah^/b
= area of 2 trl. ABH + 2 trl. DBH
= ab + a®/b. = a^ + b^. Q.E.D.

a. See Versluys, p. 8?, f*ig.
91.

Fig. 29

Ihi.r ty^-Iwg

Another Reductlo ad Absurdum
proof--see proof Sixteen above.

y

Suppose a^

+ b® > h®.

Then

A jT

C!\ti

AC^

+ p® >

b® , and

CB® + p® > a

2

/I

mmAmmmmmMy

AC

® + CB®

+ 2p® >

a® + b® > h

2

As

?30

•H

• 50

2p^

= 2 (AC

X BC) then AC^ + CB^

+

2AC

X CB

> a® +

b®, or

(AC + CB)^ >

a'

2

+ b® > t

or h®

>

+ b® >

h®, or

h^ > h^, an

ab'

-

surdity.

Similarly,

if a®

+ b® < b

L^. h^ = a^

+

b^.

Q.E.D.

a. See Versluys, p. 60, fig. 64.

Ib.iiiti-Ib.iie®

Sq. AD = (area of 4 tri*s
= 4 X tri. ABH + area of sq. KP )

= 4 X ^ab + (b - a)^ = 2ab + b^

- 2ab + a^ = a^ + b^. = a^ + b^.

a. See Math. Mo., 1858-9,
Vol. I, p. 561 , and it refers to
this proof as given by Dr. Hutton,
(Tracts, London, l8l2, 5 Vol., 8 OO)
in his History of Algebra.

Fig. 31

ALGEBRAIC PROOFS

^9

ItiiLlyL“Eau.L

p,---

7\C

ABH + sq. HE =

% \ L

r *

K *

= 2x® + 2xy + ;

• E\

^ '

+ X® = (x + y)'

I y\

Px '

= sq. of AH +

= a® + b®, Q.;

Let BH = X, and HP = y;
then AH = X + y; sq. AC = 4 trl.

fxfx + y)l . p

on AB
.2

Fig. 52 Rev. J. G. Excell, Lakewood, 0.,

July, 1928; also given hy R. A.
Bell, Cleveland, 0., Dec. 28, 1931. And it appears
in "Der Pythagoreisch Lehrsatz” (1930), by Dr. W.
Leitzmann, in Germany.

Ihir Five

✓

33a

In fig. 33a, sq. CG
= sq. AP + 4 X trl. ABH = h®
+ 2ab, ---(1)

In fig. 33b, sq. KD
= sq, KH + sq. HD + 4 x trl.
ABH = a® + b® + Sab. --(2)
But sq. CG = sq. KD, by
const *n. (l) = (2) or

•f 2ab = a^ + b^ 4- 2ab.

= a® + b®. Q.E.D,

a. See Math. Mo.,
1809, dera. 9, and there, p.
159, Vol. I, credited to Rev.
A. D. Wheeler, of Brunswick^
Me.; also see Pourrey, p. 80,
flg*s a and b; also see "Der
Pythagoreisch Lehr sat z"
(1930), by Dr. W, Leitzmann.

b. Using fig. 33a # a
second proof is: Place 4 rt.
triangles BHA, ACD, DEP and
PGB so that their legs form a

50

THE PYTHAGOREAN PROPOSITION

square whose side is HC. Then it is plain that;

1. Area of sq. HE = + 2ah +

2. Area of tri. BHA = ab/2.

3. Area of the 4 trl*s = 2ab.

4. Area of sq. AF = area of sq. HE - area of the 4
trl*s = a^ + 2ab + b^ - 2ab = a^ + b^.

5. But area of sq. AF = h^.

6. h^ = a^ 4 - b^. Q.E.D.

This proof was devised by Maurice Laisnez, a
high school boy, in the Junior -Senior High School of
South Bend, Ind., and sent to me. May l6, 1959> by
his class teacher, Wilson Thornton,

Ihirty^-Six

I

✓ %

Fig. 34

Sq. AE = sq. KD - 4ABH
= (a + b)^ - 2abj and h^ = sq.
NH + 4ABH = (b - a)^ + 2ab.
Adding, 2h^ = (a + b)^

+ (b - a)^ = 2a^ + 2b^. h^

= a^ + b^, Q.E.D.

a. See Versluys, p. 72,
fig. 78 ; also given by Saunder-
son (1682-I750); also see
Fourrey, p. 92, and A. Marre.
Also assigned to Bhaskara, the
Hindu Mathematician, 12th cen-
tury A.D. Also said to have
been known in China 1000 years
before the time of Christ.

Since trl*s ABH and CDH are similar, and CH
= b - a, then CD = h(b - a)/b, and DH = a (b - a)/b.
Draw GD. Now area of tri. CDH = J(b - a) x a(b - a)/b

= ia(b - a)Vt). —(1)

ALGKBRAIC PROOFS

51

Area of trl. DGA = JGA x aD = Jb

X b

a(b - a)

= ■|[b^ - a(b - a)]

( 2 )

- a '

Area of trl. GDC = |h ( -

- —(3)

^^8- 35 area of sq. AF = (l) + (2) + (5)

+ trl. GCP = Ja(b - a)Vb

+ - a(b - a)] + (b - a )/b + ^ab = b^, which

reduced and collected gives h^(b - a) - (b - a)a^

= (b - a)b^. h^ = a^ + b^. Q.E.D.

a. See Versluys, p. 73-^, solution 62.

b. An Arabic work of Annairizo, 900 N.O. has
a similar proof.

c. As last 5 proofs show, figures for geo-
metric proof are figures for algebraic proofs also.
Probably for each geometric proof there an alge-
braic proof.

B. — The Mean Proportional Principle

The mean proportional principle leading to
equivalency of areas of triangles and parallelograms,
is very prolific In proofs.

By rejecting all similar right triangles
other than those obtained by dropping a perpendicular
from the vertex of the right angle to the hypotenuse
of a right triangle and omitting all equations re-
sulting from the three similar right triangles thus
formed, save only equations (5), (5) and (7), as
given in proof One, we will have limited our field
greatly. But in this limited field the proofs possi-
ble are many, of which a few very interesting ones
will now be given.

In every figure imder B we will let h = the
hypotenuse, a = the shorter leg, and b = the longer
leg of the given right triangle ABH.

52

THE PYTHAGOREAN PROPOSITION

IhitilrEiflht

Since AC : AH = AH : AB, AH®

= AC X AB, and BH® = BC x BA. Then
BH® + HA® = (AC + CB)HB = AB®. .-. h®
= a + D .

Fig. 36 a. See Versluys, p. 82, fig.

92 , as given by Leonardo Pisano,
1220, In Practlca Geometriae; Wallis, Oxford, 1655;
Math. Mo. 1859> Dem. 4, and credited to Legendre’s
Geom. ; Wentworth’s New Plane Geom. , p. 158 (l895);
also Chauvenet’s Geom., I 89 I, p. 117 > Prop. X. Also
Dr. Leltzmann’s work (1930), p. 33 > fig. 34. Also
"Mathematics for the Million," (l937)> p. 155> fig.

51 (i), by Lancelot Hogben. F.R.S.

Ihirt^-Nine

Fig. 37

Extend AH and KB to L,
through C draw CD par. to AL, AG
perp. to CD, and LD par. to HB,
and extend HB to P.

BH^ =AH xHL=PHxHL= PDLH
= a^. Sq. AK = paral. HCEL
= paral. AGDL = a^ + b^. /. h^

= a^ + b^. Q.E.D.

a. See Versluys, p. 84,
fig. 94 , as given by Jules
Camlrs, I 889 In S. Revue

l^rtt

^ 1

Fig. 58

F-.n

Draw AC. Through C draw
CD par. to BA, and the perp’s AD,
HE and BP.

Trl. ABC = i sq. BG
= i rect. BD. /. sq. BG = a^

= rect. BD = sq. EP + rect. ED

= sq.
EP +

EP + (EA X ED = EH^) = sq.
EH®. But trl’s ABH and BHE

ALGEBRAIC PROOFS

55

are similar. if in tri. BHE, BH® = BE® + EH®,
then in its similar, the tri. ABH, AB® = BH® + HA®,
h® = a® + b®. Q.E.D.

a. See Sci. Am. Sup., Vol. 70, p. 582, Dec.
10, 1910 , fig. 7 — one of the 108 proofs of Arthur E
Colburn, LL.M., of Dist. of Columbia Bar.

F^rt^-jine

Const 'n obvious. Rect.

LP = 2 tri. PBH + 2 tri. ADB
= sq. HD = sq. LG + (rect. KP
= KC X CP = AL X LB = HL®)

= sq. LG + HL®.

But tri ' 3 ABH and BHL
are similar. Then as in fig. 56,
h® = a® + b®.

a. See Sci. Am. Sup., V.
70 , p. 559, one of Colburn's IO 8 .

Fig. 59

* \

Fig. 40

Construction as in fig.
? 8 . Paral. BDKA = rect. AG = AB
X BG = AB X BC = BH®. And AB
X AC = AH®. Adding BH® + AH®

= AB X BC + AB X AC = AB(AC + CB)
= AB®. h® = a® + b®. Q.E.D.

a. See Wlpper, I 88 O, p.
59, fig. 58 and there credited to
Oscar Werner, as recorded in

"Archlv. d. Math, und Phys.," Grunert, 1855; also see
Versluys, p. 64, fig. 67, and Pourrey, p. 76.

5 ^

THE PYTHAGOREAN PROPOSITION

Fig. kl

HK + sq. DB = AB^.

a. See Scl
1910. Credited to

Two squares, one on AH
const *d outwardly, the other on
HB overlapping the given triangle.

Take HD = HB and cons*t
rt. trl. CDG. Then trl*s CDH and
ABH are equal. Draw GE par. to
AB meeting GKA produced at E.

Rect. GK = rect. GA + sq.
HK = (HA = HC)HG + sq. HK = HD®

+ sq. HK.

Now GC ; DC = DC : (HO = GE)
DC® = GC X GE = rect, GK = sq.

. . h = a 4 - D .

. Am. Sup., V. 70, p. 382, Dec. 10 ,
A. E. Colburn.

AK = sq. on AB.
Through G draw GD par. to HL
and meeting PL produced at D
and draw EG.

Trl. AGE Is common to
sq. AK and rect. AD. trl.
AGE = ^ sq. AK = J rect. AD.

sq. AK = rect. AD. Rect. AD
= sq. HP + (rect. HD = sq. HO,
see argument In proof 39 ). /.

. BE = sq.

.2 . X.2

1910.

sq

= a^ + h
a. See Scl. Am. Sup., V. 70,
Credited to A. E. Colburn.

HC + HP, or
p. 382, Dec. 10 ,

b. I regard this proof, wanting ratio, as a
geometric, rather than an algebraic proof. E. S.
Loomis.

ALGEBRAIC PROOFS

55

EaLti-Flva

Fig. k 3

HG - sq. on AH. Extend
KB to M and through M draw ML
par. to HB meeting GP extended
at L and draw CM.

Trl. ACG = tri. ABH.

Trl. MAC = i rect. AL = J sq. AK.

sq. AK = rect. AL = sq. HG +
(rect. HL=ML xMH).=HA xHM
= = sq. HD) = sq. HG + sq. HD.

.*. h = a + D .

a. See Am. Sci. Sup., V.
70, p. 38?, Dec. 10, 1910. Cred-
ited to A. E. Colburn.

Ffirtjt-Six

Fig. 44

Extend KB to 0 In HE.
Through 0, and par. to HB draw
NM, making CM and ON each = to
HA. Extend GP to N, GA to L,
making AL = to AG and draw CM.

Tri. ACL = trl. 0PM
= trl. ABH, and tri. CKP
= trl. ABO.

rect. OL = sq. AK,
having polygon ALPB In common.
/. sq. AK = rect. AM = sq. HG
+ rect. HN = sq. HG + sq. HD;
see proof Porty-Pour above.

.-. h® = + b^. Q.E.D.

a. See Am. Scl. Sup.,
V. 70, p. 585. Credited to
A. E. Colburn.

56

THE PYTHAGOREAN PROPOSITION

FfirtxrSeyen

t : '' 1^1

:b

Fig. 45

Transposed sq, LE = sq. on

AB.

Draw through H, perp. to
AB, GH and produce It to meet MC
produced at P. Take HK = GB, and
through K draw LN par. and equal
to AB. Complete the transposed
sq. LE. Sq. LE = rect. DN + rect.
DL = (DK X KN = LN X KN = AB X AG
= HB^) 4* (rect. LD = paral. AP

tri. RMA
sq. LE

PCH =
. SLA.

= sq. AC) for trl.
and trl. CPR = trl
sq. AC, or h^ = a^ + h^.
a. Original with the author of this work,
Peh. 2, 1926.

= HB^ +

FjJirtz.EifliLt

Construct

trl. BHE = tri. BHC
and trl. AHP = trl.
AHC, and throu^ pts.
P. ft, and E draw the
line GHL, making PG
and EL each = AB,
and complete the
rect*s PK and ED,
and draw the lines
HD and HK.

Trl. HKA

= iAKxAF = iAB
X AC - i AH®, Trl. HBD =iBDxBE=^ABxBC=J
HB®. Whence AB x AC = AH® and AB x BC = HB®. Add-
ing, we get AB x aC + AB x BC = AB (AC + BC ) = AB®,
or AB® = BH® + HA®. h® = a® + b®.

a. Original with the author, discovered Jan.

51, 1926.

ALGEBRA.IC PROOFS

57

Fgrtyi-Hine

Construction. Drav HC, AE
and BP each perp. to AB, making each
equal to AB. Draw EC and PCD. Tri*s
^ ABH and HCD are equal and similar.

\ "I Figure PCEBHA = paral. CB

< ! j + papal. CA = CH X GB + CH X GA

[ )i I =ABxGB + ABxAG = HB® + HA®

I \ = AB(GB + AG) = AB X aB = AB®.

p a. See Math. Teacher, V. XVI,

1915 . Credited to Geo. G. Evans,

Fig. i4-7 Charleston High School, Boston, Mass.;

also Versluys, p. 64, fig. 68, and
p. 65 , fig. 69; also Journal de Mathein, I 888 ,

P. Pahre; and found in "De Vriend der Wirk, 1889 ,”
by A. E. B. Dulfer.

Fig. 47

WAl

I am giving this figure
of Cecil Hawkins as it appears
in Versluys’ work, — not reduc-
ing it to my scale of h = 1" .

Let HB’ = HB = a, and
HA* = HA = b, and draw A’B* to
, D in AB.

^ Then angle BDA’ is a rt.

angle, since trl’s BHA and B’HA’
Fig. 48 are congruent having base and

altitude of the one res’ly perp.
to base and altitude of the other.

Now trl. BHB’ + trl. AHA* = trl. BA’B* + trl.
AB’A* = trl. BAA - tri. BB’A. ^ a^ + ^ b^

= i(AB X a’D - |(AB X B’D) = ^[AB(A’B* + B’D)]

- i(AB X B’D) * i AB X a»B’ + J AB x b’D - ^ AB x b’D
= i AB X a’B’ = i h X h = i h^. .% h^ = a^ + b^.
Q.E.D.

a. See Versluys, p. 71# fig. 76, as given by
Cecil Hawkins, 1909# of England.

58

THE PYTHAGOREAN PROPOSITION

Fif

sq.

Trl. ACG = trl. ABH.
HG = quad. ABPC =

Since angle BAG = rt. angle.

/. trl. CAB = b^ = quad.

ABPC = + trl. BPC =

+ ^(b + a)(b - a). (l)

Sq. HD = sq. HD* . Trl. OD‘B
= trl. RHB. /. sq. HD^ = quad.

+ trl. ABL - trl.

= ih - J(b + a)

— (2) (1) + (2)

ib^ +

BRE'O = a
AEL.

(b - a)^

= (5 ) a -r u — ^11 T 2

Q.E.D.

Or from (l) thus: ^ (b + a ) (b - a ) = b

ih - l-a. Whence h^ = a^ + b^.

+b2=: ih^ +

= h^.

a^ + b^

a. See Versluys, p. 67, fig. 71^ as one of
Meyer ^s collection, of 1876.

I

3>l;>

Given the rt. trl. ABH.
Through B draw BD = 2BH and par.
to AH. Prom D draw perp. DE to AB.
Plnd mean prop'l between AB and AE
which Is BP. Prom A, on AH, lay
off AT = BP, Draw TE and TB, form-
ing the two similar trl * s AET and
ATB, from which AT : TB = AE : AT,
or (b - a)^ = h(h - EB), whence
EB = [h - (b - a)^]/h. — -(l)

Also EB : AH = BD : AB.

/. EB = 2ab/h. -—(2) Equating (l )
and (2) gives [h - (b - a)^]/h
= 2ab/h, whence h^ = a^ + b^.

a. Devised by the author, Peb. 28, 1926.

b. Here we Introduce the circle In finding

•if

F

IK

Fig. 50

the mean proportional.

ALGEBRAIC PROOFS

59

iiirlllLSfi

An Indirect algebraic
proof , said to be due to the
great Leibniz (l646-17l6 ) .

If (1) HA* + HB* =AB*,
then (2) HA* = AB* - HB*,
whence (?) HA* = (AB + HB)

(AB - HB),

Take BE and BC each
equal to AB, and from B as
center describe the semicircle
CA*E. Join AE and AC, and
draw BD perp. to AE. Now (4)
HE = AB + HB, and (5 ) HC = AB
- HB. (4) X ( 5 ) gives HE X HC
= HA*^, which Is true only when triangles AHC and EHA
are similar.

(6) angle CAH = angle AEH, and so (?) HC
: HA = HA : HE; since angle HAC = angle E, then angle
CAH = angle EAH. /. angle AEH + angle EAH = 90° and
angle CAH + angle EAH = 90°. .% angle EAC = 90°.
vertex A lies on the semicircle, or A coincides with
A*. EAC Is Inscribed in a semicircle and Is a rt.
angle. Since equation (l) leads through the data de-
rived from It to a rt. triangle, then starting with
such a triangle and reversing the argument we arrive
at h^ = a^ + b^.

a. See Versluys, p. 6l, fig. 65 , as given by
von Leibniz.

Eifti-Fftiir

Let CB = X, CA = y and HC
p. p® = xy; X® + p® = X® + xy

x(x + y) = a®, y® + p® = y® + xy

y(x + y) = b®. X* + 2p® + y®
a® + b®. X® + 2xy + y® = (x + y)®

a® + b®. /. h® => a® + b®. Q.E.D.

Fig. 52

60

THE PYTHAGOREAN PROPOSITION

a. This proof was sent to me by J. Adams of
The Hague, Holland. Received it March 2, 193^^
the author was not given.

Eitll-Eive

Fig. 53

Assume (l) HB‘
Draw HC perp. to AB.

4 *
Then

HA® = AB®.

(2) AC‘

+ CH® = HA®. (5) CB® + CH^" = HB^
(4) Now AB = AC + CB, so
CB + CB®

(5) AB‘

( 8 )

( 2 )

( 11 )

= AC'
+ CB'
( 7 )

+ 2AC
. But (6) HC® = AC
AB® = AC® + 2AC X CB

= AC® + 2HC®

AB = AC

+ ( 3 ) =

+ CB.

(9) AB® = AC'

V 2 _ A n2

X QB ,

+ CB®
< CB +

+ 2AC

(10) HB® + HA® = AC® + 2HC® +

+ HA®. (12) h® = a® + h

a. See Versluya, p. 62, fig. 66.

b. This proof is one of Hoffmann's, l8l8.

and

CB®.

AB® = HB®

CB®, or

. Q.E.D.

col-

lection.

C . — The Circle in Connection with the Rliht Trianile

(I ) . — Through the Use of One Circle

Prom certain Linear Relations of the Chord,
Secant and Tangent in conjunction with a right tri-
angle, or with similar related right triangles, it
may also be proven that: The square of the hypotenuse
of a riiht triangle is equal to the sun of the squares
of the other two sides.

And since the algebraic is the measure or
transliteration of the geometric square the truth by
any proof through the algebraic method Involves the
truth of the geometric method.

Furthermore these proofs through the use of
circle elements are true, not because of straight-
line properties of the circle, but because of the law
of similarity, as each proof may be reduced to the
proportionality of the homologous sides of similar
triangles, the circle being a factor only in this,
that the homologous angles are measured by equal arcs.

ALGEBRAIC PROOFS

6l

(l) The Method by Chords.

Eifty-Six

H is any pt. on the semi-
circle BHA. the trl. ABH Is a rt.

— I P triangle. Complete the sq. AP and
r draw the perp. EHC.

I BH^ = AB X BC (mean propor-

I ' tlonal)

I AH^ = AB X AC (mean propor-

i tional }

C Sq, AP = rect. BE + rect. AE = AB x BC

Fig. 54 + AB X AC = BH^ + AH^. A h^ = a^

+ b^

a. See Scl. Am. Sup., V. 70, p. 383> Dec. 10,
1910. Credited to A. E. Colburn.

b. Also by Richard A. Bell, --given to me Peb.
28, 1958. He says he produced it on Nov. I8, 1955*

Eifii

Fig. 55

3, 1910. Credited to A.

Sev^n

Take ER = ED and
Bisect HE. With Q as cen-
ter describe semicircle
AGR. Complete sq. EP .
Rect. HD = HC x HE = HA
X he = HB® = sq. HP. EG
is a mean proportional be-
tween EA and (ER = ED).

/. sq. EP = rect. AD = sq.
AC + sq. HP. But AB is a
mean prop*l between EA and
(ER = ED). EG = AB.
sq. BL = sq. AC + sq. HP.

.-. h® + a* + b®.

a. See Sci. Am.
Sup., Y. 70, p. 359, Dec.

. Colburn.

62

THE PYTHAGOREAN PROPOSITION

again,

In any circle upon
any diameter, EC In fig. 56,
take any distance from the
center less than the radius,
as BH. At H draw a chord
AD perp. to the diameter,
and join AB forming the rt.
trl. ABH.

a. Now HA X hD = HO

X HE, or b^ = (h + a)(h - a),
h® = a® + b®,

b. By joining A and
C, and E and D, two similar
rt. trl*s are formed, giv-
ing HC : HA = HD : HE, or.

= (h + a ) (h - a ) . h® = a® + b®

But by joining C and D, the trl. DHC = trl.
AHC, and since the trl. DEC Is a particular case of
One, fig. 1, as Is obvious, the above proof Is sub-
ordinate to, being but a particular case of the proof
of. One.

c. See Edwards* Geometry, p. 156, fig. 9,
and Journal of Education, I887, V. XXV", p. 404, fig.
VII.

Eiftl-Ninfe

\ /

Fig. 57

With B as center,
and radius = AB, describe
circle AEC.

Since CD Is a mean
proportional between AD and
DE, and as CD = AH, b^

= (h - a ) (h + a ) = h^ - a® .
h^ = a^ + b^.

a. See Journal of
Education, I888, Vol. XX7II,
p. 327, 21st proof; also
Heath* s Math. Monograph,

ALGEBRAIC PROOFS

65

No. 2, p. 30, 17 th of the 26 proofs there given.

h. By analysis and comparison It Is obvious,
by substituting for ABH Its equal, trl. CBD, that
above solution Is subordinate to that of Flfty-Slx .

Sixti

In any circle drav any
chord as AC perp. to any diam-
eter as BD, and jolh A and B,

B and C, and C and D, forming
the three similar rt. trl*s
ABH, CBH and DBC.

Whence AB : DB = BH
: BC, giving AB x BC = DB x BH
= (DH + HB)BH = DH x BH + BH^

= AH X HC + BH^; or h^ = a®

+ b^.

a. Fig. 58 Is closely
related to Fig. 56 .

b. For solutions see Edwards' Geom. , p. 156 ,
fig. 10, Journal of Education, I 887 , V, XXVI, p. 21,
fig. 14, Heath's Math. Monographs, No. 1, p. 26 and
Am. Math. Mo., V. Ill, p. 3 OO, solution XXI.

^ixtu-Qne

Fig. 59

« HA X HC) = BH®

Let H be the center of a
circle, and AC and BD two diameters
perp. to each other. Since HA = HB,
we have the case particular, same
as In fig. under Geometric Solu-
tions .

Proof 1 . AB X BC = BH®

+ AH X CH. AB® = HB® + HA®.
h® = a® + b®.

Proof 2 . AB X BC = BD X bH
= (BH + HD) X bH = BH® + (HD x hB
+ AH®. h® = a® + b®.

64

THE PYTHAGOREAN PROPOSITION

a. These two proofs are from Math. Mo., l859>
Vol. 2, No. 2, Dem. 20 and Dem. 21, and are applica-
tions of Prop. XXXI, Book IV, Davies Legendre, (1858),
p. 119; or Book III, p. 173, Exercise 7, Schuyler *s
Geom. , (1876), or Book III, p. I65, Prop. XXIII,
Wentworth’s New Plane Geom., (l895).

h. But it does not follow that being true
when HA = HB, it will be true when HA > or < HB. The
author.

SJLxt^-Iwg

Fig. 60

At B erect a perp. to AB and
prolong AH to C, and BH to D. BH
= HD. Now AB® = AH x AC = AH{AH + HC)
= AH® + (AH X HC = HB®) = AH® + HB®.
h® = a® + b®. Q.E.D.

a. See Versluys, p. 92, fig.

105.

Sixt^-Three

Prom the figure it
is evident that AH x HD
= HC X he, or b^ = (h + a)

(h - a) = h® - a®. /. h®

= a® + b®. Q.E.D.

a. See Versluys,
p. 92, fig. 106, and credit-
ed to Wm. W. Rupert, I9OO.

Fig. 61

ALGEBRAIC PROOFS

65

Sixtjr-Fgjir

With CB as radius
describe semicircle BHA cut-
ting HL at K and AL at M.

Arc BH = arc KM. BN = NQ
= AO = MR and KB = KAj also
arc BHK = arc AMR = MKH = 90®.
So trl's BRK and KLA are con-
gruent. HK = HL - KL = HA
- OA. Now HL : KL = HA : OA.
Fig. 62 So HL - KL : HL = HA - OA :HA,

or (HL - KL) T HL = (HA - OA )

T HA = (b - a )/b . KQ = (HK -f NL )LP = [ (b - a ) t b]

X Jb = i(b - a).

Now trl. KLA = trl. HLA - trl. AHK = ^b®

- ib X J(b - a) = iba = ^ trl. ABH, or trl. ABH

= trl. BKR + trl. KLA, whence trap. LABR - trl. ABH
= trap. LABR- (trl. BKR + trl. KLA) = trap. LABR

- (trl. HBR + trl. HAL) = trap. LABR - trl. ABK.

trl. ABK = trl. HBR + trl. HAL; or 4 trl. ABK = 4 trL
HBR + 4 trl. HAL. /. = a^ + b^. Q.E.D.

a. See Versluys, p. 95 ^ f*lg. 107; and found
In Journal de Matheln, 1897^ credited to Brand.

(10/25, '55, 9 P. m. E. S. L. ).

Sixt)[,-Fi,ve

Fig. 63

Geom. , 6th Ed'n,

The construction Is obvious.
From the similar triangles HDA and
HBC, we have HD : HB = AD : CB, or
HD X CB = HB X AD. ---(l)

In like manner, from the
similar triangles DHB and AHC, HD
X AC = AH X DB. ---(2) Adding (l )
and (2), HD>‘AB = HB xAD+AH
X DB. —(5). h® = a® + b®.

a. See Halsted's Elementary
1895 for Eq. (3), p. 202; Edwards'

Geom. , p. 158, fig. 17; Am. Math. Mo., V. IV, p. 11.

66

THE PYTHAGOREAN PROPOSITION

b. Its first appearance in print. It seems,
was In Runkle*s Math. Mo., 1859, and by Runkle cred-
ited to C. M. Raub, of Allentown, Pa.

c. May not a different solution be obtained
from other proportions from these same triangles?

Sixt3t-Slx

BC

P.

Ptolemy’s Theorem (A.D. 87-
l68). If ABCD Is any cyclic (in-
scribed) quadrilateral, then AD x
+ AB X CD = AC X bD.

As appears In Wentworth’s
Geometry, revised edition (l895),

176, Theorem 238. Draw DE making
Z.CDE = ZADB. Then the trl’s ABD and
CDE are similar; also the trl’s BCD
and ADE are similar. Prom these pairs of similar
triangles It follows that AC x BD = AD x BC + DC x AB.
(For full demonstration, see Teacher’s Edition of
Plane and Solid Geometry (1912), by Geo. Wentworth
and David E. Smith, p. I90, Proof 11.)

In case the quad. ABCD be-
‘v comes a rectangle then AC = BD, BC
= AD and AB = CD. So AC® = BC®

+ AD®, or c® = a® + b®. a special
case of Ptolemy’s Theorem gives a
proof of the Pyth. Theorem.

a. As formulated by the
author. Also see ”A Companion to
Elementary School Mathematics (1924),
by P. C. Boon, B.A., p. 107, proof 10.

45 /

r ^

rig. 65

S j_xt^-Seven

Circumscribe about trl. ABH circle BHA. Draw
AD = DB. Join HD. Draw CG perp. to HD at H, and AC
and BG each perp. to CG; also AE and BP perp. to HD.

Quad’s CE and PG are squares. Trl’s HDE and

ALgBBRAIC PROOFS

67

Fig. 66

DBF are congruent. AE = DP = EH
= AC. HD = HP + PD = BG + AC. Quad.
ADBH = ^HD(BP + AE) = ^HD x CG.

Quad. ABGC = i (AC + BG) x cG = ^HD
X CG. tri. ADB = trl. AHC + trl.
HBG. 4 tri. ADB = 4 trl. AHC + 4
trl. HBG. .*. h® = a® + b®. Q.E.D.

a. See E. Pourrey's C. Geom.,
1907 j credited to Plton-Bressant;
see Versluys, p. 90, fig. 105.

b. See fig. 333 for Geom. Proof — so-called.

Siity-Eiaht

Fig. 67

Construction same as in
fig. 66, for points C, D and G.
Join DG. Prom H draw HE perp. to
AB, and join EG and ED. Prom G
draw GK perp. to HE and GP perp.
to AB, and extend AB to P. KP is
a square, with diag. GE. angle
BEG = angle EBD = 45°. GE and
BD are parallel. Trl. BDG = trl.
BDE. -—(1) Trl. BGH = trl. BGD.
— (2) .-. (1) = (2), or trl. BGH

= trl. BDE. Also trl. HCA = trl. ADE. tri. BGH
+ tri. HCA = trl. ADB. So 4 trl. ADB = 4 trl. BHG
+ 4 trl. HCA. h* = a® + b®. Q.E.D.

a. See Versluys, p. 91, fig. 104, and credit-
ed also to Plton-Bressant, as found in E. Pourrey's
Geom., 1907^ p. ISf IX.

b. See fig. 354 of Geom. Proofs.

BDE. Also trl,
HCA = trl. ADB.

trl. BGH

Pourrey ' s

Sixti-aina

In fig. 63 above it is obvious that AB x bH
= AH X DB + AD X BH. /. AB^ = HA® + HB®. h® = a®

+ b®.

a. See Math. Mo., 1859, by Runkle, Vol. II,
No. 2, Dem. 22, fig. 11.

68

THE PYTHAGOREAN PROPOSITION

b. This Is a particular case of Prop, XXXIII,
Book IV, p. 121, Davies Legendre (1858) vhlch is Ex-
ercise 10, in Schuyler *s Geom. (1876), Book III, p.
173, or Exercise 238, Wentworth’s New Plane Geom.
(1895), Book III, p. 176.

Seventjc.

On any diameter as AE
= 2AH, const, rt. trl. ABH, and
produce the sides to chords.

Draw ED. Prom the slm. trl’s
ABH and AED, AB : AE = AH : AD,
or h : b + HE = b : h + BD.
h (h + BD ) = b (b + HE = + b

X HE = b® + HP X HC = b® + HC^.

(1) Now conceive AD to re-

volve on A as a center until D
coincides with C, when AB = AD
= AC = h, BD = 0, and HB = HC
= a. Substituting in (l) we have h^ = a^ + b^.

a. This is the solution of G. I, Hopkins of
Manchester, N.H. See his Plane Geom,, p. 92, art.
427; also see Jour, of Ed., I888, V. XXVII, p. 327,
16th prob. Also Heath’s Math. Monographs, No. 2,
p. 28, proof XV.

b. Special case. When H coincides with 0 we
get (1) BC = (b + c)(b - a)/h, and (2) BC = 2bVh - h.

Equating, h® = a® + b^.

c. See Am. Math. Mo., V. Ill, p. 500.

(2) The Method by Secants.

Sevent^-Jne

With H as center and HB as radius describe
the circle EBD.

The secants and their external segments bring
reciprocally proportional, we have, AD : AB = AP : AE,

ALGEBRAIC PROOFS

or b + a : h = (h - 2CB = h

: b - a, whence h = a + b .

/ a. In case b = a, the

/ \ points A, E and P coincide and the

! ^ proof still holds; for substituting

\ ^ ^ above prop'n reduces to

h® - 2a® = 0 ; h® = 2 a® as It
p jf should.

Fig. 69 b. By joining E and B, and

P and D, the similar triangles upon
which the above rests are formed.

Seven ty-Twg

...... With H as center and HB as

radius describe circle FBD, and draw
/ » HE and HC to middle of EB.

; • AE X AB = AP X AD, or

; (AD - 2BC )AB = (AH - HB ) (AH + HB ) .

Jl 1 X ' AB* - 2BC X aB = AH® - HB® . And

as BC ; BH = BH ; AB, then BC x aB
** = HB®, or 2BC x aB = 2BH®. So AB®

Fig. 70 - 2BH® = AH® - BH®. AB® = HB®

+ HA®. h® = a® + o®. Q.E.D.
a. Math. Mo., Vol. II, No. 2, Dem. 25, fig. 2.
Derived from: Prop. XXIX, Book IV, p. II 8 , Davies
Legendre ( 1858 ); Prop. XXXIII, Book III, p. 171,
Schuyler's Geometry ( 1876 ); Prop. XXI, Book III, p.
163 , Wentworth's New Plane Geom. (1895)*

AE : AH = AH : AD.

AH® = AE * AD = AE(AB x bH)

= AE X AB + AE X BH. So AH®
+ BH® =AExAB+AExHB
+ HB® = AE X aB + HB(AE+BH)
= AB(AE + BH) = AB®. h®

= a® + b®. Q.E.D.

a. See Math. Mo.,

Pig. 71

70

THE PYTHAGOREAN PROPOSITION

(l859)> Vol. II, No, 2, Dem. 26, p. 15; derived from
Prop. XXX, p. 119, Davis Legendre; Schuyler *s Geom. ,
Book III, Prop. XXXII, Cor. p. 172 (1876); Went-
worth* s Geom., Book III, Prop. XXII, p. l64. It Is
credited to C. J. Kemper, Harrisonburg, Va., and Prof.
Charles A. Yoiing (l859)# at Hudson, 0. Also foxind
In Pourrey*s collection, p. 93# as given by J. J. I.
Hoffmann, 1821.

In fig. 72, E will fall be-
tween A and P at P, or between P
and B, as HB Is less than, equal to,
or greater than HE. Hence there
are three cases; but Investigation
of one case — when it falls at middle
point of AB — Is sufficient.

Join L and B, and P and C,
making the two similar triangles

APC and ALB; whence h : b + a = b
2 2

- a : AP; AP = • — (l)

Join P and G, and B and D making the two sim-
ilar trl*s PGE and BDE, whence Jh : a - ^h = a + Jh
a^ - ih^

: PE, whence PE =
(2) gives ih =

ih

+

h

— (2 ) . Adding (1 ) and

; whence h^ = a^ + b^.

a. The above solution Is given by Krueger,

In "Aumerkungen uber Hrn. geh. R. Wolf’s Auszug aus
der Greometrle," 17^6. Also see Jury Wlpper, p. 4l,
fig. 42, and Am. Math. Mo., V. IV, p. 11.

b. When G falls midway between P and B, then
fig. 72 becomes fig. 69. Therefore cases 69 and 72
are closely related.

ALC^RAIC PROOFS

71

§fiV£at)tiFiV£

In fig. 73a, take HF
= HB. With B as center, and
BP as radius describe semi-
circle DEG, G being the pt.
where the circle Intersects
AB. Produce AB to D, and
draw PG, PB, BE to AH pro-
duced, and DE, forming the
similar trl*s AGP and AED,
from which (AG = x ) : (AP = y )

= (AE = y + 2PH ) : (AD = x

+ 2BG) = y 4- 2z : x + 2r

whence x^ + 2rx = y^ + 2yz.
— ( 1 ).

But If, see fig. 75b,
HA = HB, (sq. GE = h^) = (sq.

HB = a^) + (4 trl. AHG = sq.

J.XA — Kf J f XX — C* ~ + b^; then, (see fig. 75a)

when BP = BG, we will have BG^ = HB^ + HP®, or r®

= z® + z®, (since z = PH). (2).

(l) + (2) = (5) X® + 2rx 4- r® = y® + 2yz 4- z®

4* z® or (4) (x 4- r)® = (y 4* z)® + z®. (5) h® = a®

4- b®, since x4-r = AB«h, y+z = AH = b, and
z = HB = a.

a. See Jury Wlpper, p. 56, where Wlpper also
credits It to Joh. Hoffmann. See also Wlpper, p. 57^
fig. 5^, for another statement of same proof; and
Pourrey, p. Sk, for Hoffmann’s proof.

Sftveaii-Sijc

In fig. 7^ In the circle whose center Is 0,
and whose diameter Is AB, erect the perp. DO, join D
to A and B, produce DA to P, making AP = AH, and pro-
duce HB to G making BG = BD, thus forming the two
Isosceles trl’s PHA and DGB; also the two isosceles
trl’s ARD and BHS. As angle DAH = 2 angle at P, and
angle HBD » 2 angle at G, and as angle DAH and angle

72

THE PYTHAGOREAN PROPOSITION

HBD are measured by same arc
HD, then angle at P = angle
at G. arc AP = arc QB.

And as angles ADR
and BHS have same measure, J
of arc APQ, and i of arc BQP,
respectively, then tri's ARD
and BHS are similar, R Is the
Fig. 7^ Intersection of AH and DG,

and S the Intersection of BD
and HP. Now since trl'a PSD and GHR are similar, be-
ing equiangular, we have, DS : DP = HR : HG. .-. DS
: (DA + AP) = HR : (hB + BG).

DS : (DA + AH) = HR : (HB + BD),

DS : (2BR + RH) = HR : (2BS + SD),

.-. (l) DS® + 2DS X BS = HR® + 2HR x BR.

And (2) HA® = (HR + RA)® = HR® + 2HR x ra + RA® = HR®
+ 2HR X RA + AD®

(3) HB® = BS® = (BD - DS)® = BD® - 2BD x ps + 038

= AD® - (2BD X DS - DS® ) = AD® - 2 (BS + SD )DS + DS®

= AD® - 2BS X SD _ 2DS® + DS® = AD® - 2BS x DS

- DS® = AD® - (2BS X DS - DS® )

(2) + (3) = (4) HB® + HA® = 2AD®. But as In proof,

Fig. 73b, we found, (eq. 2), r® = z® + z® = 2z®.

.-. 2AD® (In fig. 74) = AB®. .-. h® = a® + b®.

a. See Jury Nipper, p. 44, fig. 43, and there

credited to Joh. Hoffmann, one of his 32 solutions.

In fig. 75, let BCA be any triangle, and let
AD, BE and CP be the three perpendiculars from the
three vertlcles. A, B and C, to the three sides, BC,
CA and AB, respectively. Upon AB, BC and CA as diam-
eters describe circumferences, and since the angles
ADC, BEC and CPA are rt. angles, the circumferences
pass through the points D and E, P and E, and P and
D, respectively.

ALGEBRAIC PROOFS

75

74

THE PYTHAGOREAN PROPOSITION

measurement of a parallelogram, (see fig. 308, this
text), we have (l ) sq. AK = (BP x AP - AM x aP )

= a(a + x). But, In trl. MCA, ON Is a mean propor-
tional between AN and NM. (?.) b^ = ax. (l) - (2)

= (;5) h^ - b^ = a^ + ax - ax = a^. h^ = a^ + b^.
Q.E.D.

a. This proof is No. 99 of A. R. Colburn* s
108 solutions, being devised Nov. 1, 1922.

(3) The Method by Tangents

1st. — The Hyvotenuse as a Tangent

SeY®QLtY"!liD.®

Draw HC perp. to AB, and
with H as a center and HC as a radi-
us describe circle GDEP.

Prom the similar trl*s ACG
and AEC, AC : AE = AG : AC, or
AO : b + r = b - r : AC; (l ) AC®

= b® - r®. Prom the similar trl's
CBD and BPC, we get (2) CB® = a®- r®.
Prom the similar rt. trl*s BCH and
HCA, we get (3) BC x AC = r®,

(4) 2BC X AC = 2r®. (l ) + (2) + (4) gives (5) AC®

+ 2AC X BC + BC® = a® + b® = (AC + BC)® = AB®. h®

= a® + b®.

a. See Am. Math. Mo., V. Ill, p. 300.

Pig. 78

- 2br. But (2)

+ a® = a® + b® -

0, the center of the circle,
lies on the bisector of angle B, and
on AH.

With the construction com-
pleted, from the similar trl*s ACD
and AHC, we get, calling OC = r,

(AC = h - a ) : (AH = b ) = (AD = b - 2r )

: (AC = h - a). (1) (h - a)® = b®

a® = a®. (1) + (2) = (3) (h - a)®

2br, or (h - a)® + 2br + a® = a®+ b®.

ALGEBBRAIC PROOFS

75

Also (AC = h - a) ; (AH = b) = (OC = OH = r) : (IIB = a),
whence

(4) (h - a)a = br.

/. (5) (h - a)^ + 2(h - a)r 4- a^ = a^ + b^

/. (6) = a^ + b^.

Or, in (3) above, expand and factor gives

(7 ) h^ - 2a (h - a ) = a^ + b^ - 2br. Sub. for

a (h - a) its equal, see (4) above, and collect,
we have

(8) h^ = a^ + b^

a. See Am. Math. Mo., V. IV, p. 8l.

2nd. — The Hyvotenuse a Secant Vhich Pass-
es Through the Center of the Circle
and One or Both Leis Tangents

liatiil-ftQe

Having HB, the shorter leg,
a tangent at C, any convenient pt.
on HB, the construction is evident.

From the similar trl * s BCE
and BDC, we get BC : BD = BE ; BC,
whence BC^ = BD x BE = (BO + OD)BE
= (BO + OC)BE.---(l) Prom similar
trl' 3 OBC and ABH, we get OB : AB

= 00 : AH, whence - Z >

' h b '

= ^ . — (2) BC : BH = OC : AH, whence BC = -—(3)

Substituting (2) and (3) in (l), gives.

^ + r)BE. . - OC) -

(if) whence h^ = a^ + b^. Q.E.D.

a. Special case is: when, in Pig. 79$ 0 co-
incides with A, as in Pig. 80.

76

THE PYTHAGOREAN PROPOSITION

llaiiiY-lMs

.r-'-'-l

With A as center and
AH as radius, describe the
semicircle BHD.

Prom the similar tri-
angles BHC and BDH, we get,
h-b:a=a:h+b, whence
directly h^ = a^ + b^,

\ J a. This case is found

in: Heath* s Math. Monographs,

No. 1, p. 22, proof VII; Hop-

Fig. 80 kins* Plane Geom. , p. 92, fig.

IX; Journal of Education,

1887 , V. XXVI, p. 21, fig. VIII; Am. Math. Mo., V.
Ill, p. 229; Jury Wlpper, I 88 O, p. 39, fig. 39, where
he says it is found in Hubert *s Elements of Algebra,
Wurceb, 1792, also in Wipper, p. 40, fig. 40, as one
of Joh. Hoffmann* s 32 proofs. Also by Richardson in
Runkle*s Mathematical (Journal) Monthly, No. 11, 1859
--one of Richardson* s 28 proofs; Versluys, p. 89 ,

fig. 99 .

b. Many persons. Independent of above sources,
have found this proof.

c. When 0, in fig. 80 , is the middle pt. of
AB, it becomes a special case of fig. 79.

Assume HB < HA, and employ
tang. HC and secant HE, whence HC^

= HE xhD»*=AD xAE = AG xaP = BP
X BG = BC^. Now employing like argu-
ment as in proof Eighty-One we get
h^ = a^ + b^.

** — a. When 0 is the middle

Fig. 81 point of AB, and HB = HA, then HB

and HA are tangents, and AG = BP,
secants, the argument is same as (c), proof Eighty -
Two , by applying theory of limits.

b. When 0 is any pt. in AB, and the two legs

ALGEBRAIC PROOFS

77

are tangents. This Is only another form of fig. 79
above, the general case. But as the general case
gives, see proof, case above, h^ = a^ + b^, therefore
the special must be true, whence In this case (c )
h^ = a^ + b^. Or If a proof 'bj explicit argiainent Is
desired, proceed as In fig. 79.

p. 80;

By proving the general case,
as In fig. 79, and then showing that
some case Is only a particular of the
general, and therefore true immedi-
ately, Is here contrasted with the
following long and complex solution
of the assumed particular case.

The following solution Is
given In The Am. Math. Mo., V. IV,

"Draw OD perp. to AB. Then, AT^ = AE x AP = AO^ - EO^
= AO® - TH®.---(1)

BP® = BP X BE = BO® - PO® = BO® - HP®.— (2)

Now, AO : OT = AD : OD;

AO X OD = OT X ad.

And, since OD = OB, OT = TH = HP, and AD = AT + TD

= AT + BP.

AT X TH + HP X BP = AO X OB.-— (5)

Adding (1), (2), and 2 x (3),

AT® + BP® + 2AT X TH + 2HP X BP = AO® - TH® + BO®

- HP® + 2A0 X OBj

AT® + 2AT X tH + PH® + BP® + 2BP x HP + HP® = AO®

+ 2A0 X OB + BO®.

(AT + TH)® + (BP + CP)® = (AO + OB)®.

AH® + BH® = AB®." Q.E.D.
h® = a® + b®.

3rd. --The Hypotenuse a Secant Hot Pass-
ini Throuih the Center of the Cir-
cle, and Both Lets Tanients

ALGEBRAIC PROOFS

79

(8) h(a + b - h - r) + ar -f br = ab.

(l) = (9) I* = (a + b - h - r). (9) in (8) gives

(10) hr + ar + br = ab.

(11) But hr + ar + br = 2 area trl. ABC.

(12) And ab = 2 area trl. ABC.

(15 ) hr + ar + br = ab » hr + r (a + b ) = hr
+ r (h + 2r )

/. (l4) 4hr + 4r® = 2ab.

/.the supposition in (4) is true.

(15) h^ = a^ + b^. Q.E.D.

a. This solution was devised by the author
Dec. 13, 1901, before receiving Vol. VIII, I9OI, p.
258, Am. Math. Mo., where a like solution is given;
also see Pourrey, p. 94, where credited.

b. By drawing a line OC, in fig. 84, we have
the geom. fig. from which. May, I89I, Dr. L. A. Bauer
of Carnegie Institute, Wash., D.C., deduced a proof
through the equations

(1) Area of trl ABH = Jr(h -f a + b), and

(2) HD + HE = a + b -h. See pamphlet: On
Rational Right-Angled Triangles, Aug., 1912, by
Artemus Martin for the Bauer proof. In same pamphlet
is still another proof attributed to Lucius Brown of
Hudson, Mass.

c. See 01ney*s Elements of Geometry, Unlversl
ty Edition, p. 312, art. 971> or Schuyler's Elements
of Geometry, p. 353^ exercise 4; also Am. Math. Mo.,
V. IV, p. 12, proof XXVI; also Versluys, p. 90, fig.
102; also Grunert's Archlv. der Matheln, and Physik,
1851 , credited to MSllmann.

d. Remark , — By ingenious devices, some if not
all, of these in which the circle has been employed
can be proved without the use of the circle — not
nearly so easily perhaps, but proved. The figure,
without the circle, would suggest the device to be
employed. By so doing new proofs may be discovered.

80

THE PYTHAGOREAN PROPOSITION

Complete rect. HG. Produce
DO to P and EO to K. Designate AC
= AE by p, BD = BO by q and HE = HD

Then a=q+r, b=p+r,
and h = p + q. TrL PMA = trl. OMC
and trl. COL = trl. KLB.

Fig. 85 trl. AGB = rect. PGKO

= trl. ABH = J rect. HG. Rect. PGKO
= rect. APOE 4 - sq. ED + rect. OKBD.

So pq = pr + r^ + qr.
whence 2pq = 2qr + 2r^ + 2pr.

But p^ + a^ = p^ + q^.

/. p^ + 2pq + q^ = (q^ + 2qr + r^ ) + (p^ + 2pr + r^ )
or (p + q)^ = (q + r)^ + (p + r)^

/. h^ = a^ + b^,

a. Sent to me by J. Adams, from The Hague,
and credited to J. P. Vaes, XIII, 4 (1917).

(II ) , — Through the Use of Two Circles .

liatiti-Eiaht

Construction, Upon the
legs of the rt. trl, ABH, as diam-
eters, construct circles and draw
HO, forming three similar rt. trl*s
ABH, HBC and HAG.

Whence h : b = b : AC. /. hAC
= b=.— (1)

Also h: a = a : BC. hBC
= a^— (2)

(1) + (2) = (5) h® = a® + Q.E.D.

a. Another form Is:

( 1 ) HA^ = HC X AB. (2) = BC x aB.

AddlJig, (3) AH® + BH® = AC x aB + BC x aB
= AB(AC + BC) = AB®. h® = a® + b®.

ALGEBRAIC PROOFS

81

Id. See Edwards’ Elements of Geom. , p. l6l,
fig. 34 and Am. Math. Mo., V. IV, p. 11; Math. Mo.
(1859), vol. II, No. 2, Dem. 27, fig. 13; Davies
Legendre, I 858 , Book IV, Prop. XXX, p. 119; Schuyler’s
Geom. ( 1876 ), Book III, Prop. XXXIII, cor., p. 172;
Wentworth’s New Plane Geom. (l895). Book III, Prop.
XXII, p. l64, from each of said Propositions, the
above proof Eighty-Eight may be derived.

Iiab.tx-S.LQ.®

with the legs
of the rt. trl. ABH as
radii describe circum-
ferences, and extend AB
to C and P. Draw HC,
HD, HE and HP. Prom
the similar trl’s AHP
and HDH,

AP ; AH = AH : AD
b® = AP X AD.— -( 1 )

Prom the similar trl’s CHB and HEB,

CB : HB = HB : BE. a® = CB x BE. (2)

(1) + (2) = (3) a® + b® = CB X be + AP X ad
= (h + b)(h-b)+ (h + a)(h-a)

= h® - b® + h® - a®j
(4) 2h® = 2a® + 2b®. h® = a® + b®.
a. Am. Math. Mo., V. IV, p. 12; also on p. 12
Is a proof by Richardson. But It Is much more dif-
ficult than the above method.

iliaaix

Por proof Ninety use fig. 87.

AH® = AD(AB + BH). — (l) BH® = BE (BA + AH).— (2)

(1) + (2) = (3) BH® + AH® = BH(BA + AH) + AD(AB + BH)
=BHxBA+BExaH+ADxHB+ADxBH
= HB(BE+AD)+ADxBH+BExAH + BExAB-BExAB

82

THE PYTHAGOREAN PROPOSITION

= AB(BE + AD) + AD x BH + BE (AH + AB ) - BE x aB
= AB (BE + AD) + AD x bH + BE (AH + AE -f* BE) - BE x aB
= AB(BE + AD) + AD x bH + BE (BE + 2AH) - BE x aB

= AB (BE + AD ) + AD x bH + BE^ + 2BE x aH - BE x aB

= AB(BE + AD) + AD X bH -f BE^ + 2BE x aE - BE(AD + BD)

= AB (be + AD ) + AD x bH + BE^ + 2BE x aE - BE x aD

- BE X BD

= AB(BE + AD) + AD X BH + BE (BE + 2AE) - BE (AD + BD )

= AB(BE + AD) + AD x BH + BE (AB + AH) - BE (AD + BD)

= AB (BE + AD ) + AD x BH + (BE x BC = BH^ = BD^ )

- BE(AD + BD)

= AB(BE + AD) + (AD + BD)(BD - BE)

= AB (BE + AD ) + AB x DE = AB (BE + AD + DE )

= AB X AB = AB^. /. Q.E.D.

a. See Math. Mo. (l859)> Vol. II, No. 2, Dem.
28, fig. 13 — derived from Prop. XXX, Book IV, p. 119^
Davies Legendre, I858; also Am. Math. Mo., Vol, IV,
p. 12, proof XXV.

Ninet^-gns

For proof Ninety-One use fig. 87. This proof
Is known as the "Harmonic Proportion Proof.”

From the similar trl*s AHF and ADH,

whence

or

and

or

AH : AD = AF : AH, or AC : AD = AF : AE
AC+AD :AF + AE = AD :AE
CD : OF = AD ; AE,

AC - AD = AF - AE = AD : AE,

DE : EF = AD : AE.

/. CD : CF = DC : EF.

or (h + b- a):(h + b + a)=(a-h + b):(a + h + b)

.*. by expanding and collecting, we get

V ,2 . 1^2

h = a 4- b .

a. See Olney’s Elements of Geom., University
Ed*n, p. 312, art. 971^ or Schuyler’s Elements of
Geom., p. 353^ Exercise 4; also Am. Math. Mo., V. IV,
p. 12, proof XXVI.

ALGEBRAIC PROOFS

83

Z ). — Ratio of Areas

As in the three preceding divisions, so here
in D we must rest our proofs on similar rt. triangles.

Draw HO perp. to AB, form-
ing the three similar triangles ABH,
ARC and HBC, and denote AB = h, HB
= a, HA = h, AC = X, OB = y and HO

Since similar surfaces are
proportional to the squares of their
homologous dimensions, therefore,

[i(x + y)z -5- iyz = h^ -f a^] = [ ^yz + ^xz = a^ -f b^]
= [i(x + y)z -i- iyz = (a^ + b^)a^]
h^ + a® = (a^ + b^) -f a^
h^ = a^ + b^.

a. See Jury Wlpper, l880, p. 38, fig. 36 as
found in Elements of Geometry of Bezout; Pourrey,
p. 91, as in Wallis* Treatise of Algebra, (Oxford),
1685; p. 93 of Cours de Mathematlques, Paris, I768.
Also Heath’s Math. Monographs, No. 2, p. 29, proof
XVI; Journal of Education,' I888, V. XXVII, p. 327,
19th proof, where it is credited to L. J. Bullard,
of Manchester, N.H.

Hinaty-Ihr^e

As the trl’s ACH, HOB and
ABH are similar, then trl. HAC : trl.
BHC : trl. ABH = AH^ : BH^ : AB^,
and so trl. AHC + trl. BHC : trl.

ABH = AH^ + BH^ : AB^. Now trl. AHC
Fig. 89 + trl. BHC : trl. ABH = 1 . AB^

= BH^ + AH^. h^ = a^ + b®. Q.E.D.
a. See Versluys, p. 82, proof 77, where cred-
ited to Bezout, 1768; also Math. Mo., 1859, Vol. II,
Dem. 5, P- ^5; also credited to Oliver; the School

N

Fig. 88

84

THE PYTHAGOREAN PROPOSITION

Visitor, Vol. 20, p. 167, says Pythagoras gave this
proof — hut no documentary evidence.

Also Stanley Jashemskl a school boy, age 19,
of So. High School, Youngstown, 0., In 1934, sent me
same proof, as an original discovery on his part.

b. Other proportions than the explicit one
as given above may be deduced, and so other sym-
bolized proofs, from same figure, are derivable --
see Versluys, p. 83 , proof 78.

Ninetj'-Fjjur

Trl*3 ABH and ABH* are con-
gruent; also trl*s AHL and AHP : also
trl's BKH and BPH. Trl. ABH = trl.
BHP + trl. HAP = trl. BKH + trl. AHL.

trl. ABH : trl. BKH : trl. AHL = h^
: a^ : b^, and so trl. ABH : (trl.

BKH + trl. AHL) =h^ : a^ + b^, or
1 = h^ -f (a^ + b^). h^ = a^ + b^.
Q.E.D.

a. See Versluys, p. 84, fig.
93, where It Is attributed to Dr. H. A. Naber, I908.
Also see Dr. Leltzmann’s work, 1930 ed*n, p. 55, fig.
35.

Ninet^-F j^ve

Complete the paral. HC, and
the rect. AE, thus forming the simi-
lar trl’s BHE, HAD and BAG. Denote
the areas of these trl’s by x, y and
z respectively.

Then z : y : x = h^ : a^ : b^.

But It Is obvious that z

Fig. 91

X + y.

h^ =

a + b ,

a. Original with the author, March 26, 1926,

10 p.m.

ALGEBRAIC PROOFS

85

Ninet^-S j,x

Draw HL perp. to AB.
Since the trl*s ABH, AHL, and
HBL are similar, so also the
squares AK, BE and HG, and
since similar polygons are to
each other as the squares of
their homologous dimensions,
we have

tri. ABH : trl. HBL : tri. AHL
= h® : a® : b®

= sq. AK : sq. BE : sq. HG.

But tri. ABH = trl. HBL + trl.
AHL. sq. AK = sq. BE + sq.
HG, h® = a® + b®.

a. Devised by the author, July 1, I 9 OI, and
afterwards, Jan. 13, 195^# found in Pourrey's Curio
Geom., p. 91^ where credited to R. P. Lamy, I 685 .

!li! 16 llr.Seven

Use fig. 92 and fig. 1.

Since, by equation ( 5 ), see fig. 1, Proof
One , BH^ = BA x bl = rect. LK, and in like manner,

AH^ = AB X aL = rect. AC, therefore sq. AK = rect.

LK + rect. AC = sq. BE + sq. HG.

h® = a® + b®. Q.E.D.

a. Devised by the author July 2, I 9 OI.

b. This principle of "mean proportional" can
be made use of in many of the here-ln-after figures
among the Geometric Proofs, thus giving variations

as to the proof of said figures. Also many other fig-
ures may be constructed based upon the use of the
"mean proportional" relation; hence all such proofs,
since they result from an algebraic relationship of
corresponding lines of similar triangles, must be
classed as algebraic proofs.

86

THE PYTHAGOREAN PROPOSITION

B. — Aliebralc Proof, Throuih Theory of Limits

Ulnsix-Eiflht

\

\

\

A

✓ I

\n'' '

N 3 ^ *

\L' '

V :

cUl'-l.-'-vX

Fig. 95

(See Plato* 3 Dialogues,
Edition of 1885, Jovett
and Sons. )

The so-called Pytha-
gorean Theorem, In Its simp-
lest form Is that In which
the two legs are equal. The
great Socrates (h. 500 B.C.),
by drawing replies from a
slave, using his staff as a
pointer and a figure on Lhe
pavement (see fig. 95) as a
model, made him (the slave)
see that the equal triangles
in the squares on HB and HA
were just as many as like
equal tri*s in the sq. on AB,
as is evident by Inspection.
Meno, Vol. I, pp. 256-260,

* s translation, Chas . Scribner

a. Omitting the lines AK, CB, BE and PA,
which eliminates the numbered triangles, there re-
mains the figure which. In Free Masonry, is called
the Classic Form, the form usually found on the mas-
ter* s carpet.

b. The following rule is credited to Pytha-
goras. Let n be any odd number, the short side;
square It, and from this square subtract 1; divide
the remainder by 2, which gives the median side; add
1 to this quotient, and this sum Is the hypotenuse;
e.g., 5 = short side; 5^ - 1 = 24; 24 + 2 = 12, the
median side; 12 + 1 = 15 the hypotenuse. See said
Rule of Pythagoras, above, on p. 19.

Starting with fig. 93# and decreasing the
length of AH, which necessarily Increases the length

ALGEBRAIC PROOFS

87

^r\ j /

I N

' JJ / X
! \

£1^—

Fig. 9*+a

'A**
•*^• 5 ' 1

> I Ki I

’ 4i »*s

of AH, which necessarily in-
creases the length of HB,
since AB remains constant, we
> decrease the sq, HD and in-
/ crease the sq. HC (see fig.

, / 94a).

Now we are to prove
that the Siam of the two vari-
( able squares, sq. HD and sq.

• HC will equal the constant

J sq. HP.

Ip We have, fig. 9^a,

h^ = a^ 4 . b^ — (1)

But let side AH, fig.
93, be diminished as by x,
thus giving AH, fig. 9^a, or bet-
ter, PD, fig. 9^b, and let DK be
Increased by y, as determined by
the hypotenuse h remaining con-
stant .

Now, fig. 9^t», when a = b,
a^ + b^ = 2 area of sq. DP. And
3 when a < b, we have (a - x)^

= area of sq. DN, and (b + y)^

= area of sq. DR.

Also c^ - (b + y)^

= (a - x)^ = area of MABCLR, or

(a - x)^ + (b + y)^ = c^. (2)

Is this true? Suppose it is;
then, after reducing (2) - (l)

= ( 3 ) - 2ax + x^ + 2by + y^ = 0,
or ( 4 ) 2ax - x^ = 2by 4- y^, which
shows that the area by which
(a^ = sq. DP) is diminished = the
is increased. See graph 9^t). the

Fig. 94b = area of sq. DR.

Also c^ -

DA = AB = c = (a - x)^ = aref

]2E = DK = a = b (a - X )^ + (b 4- 3

DF = a - X Is this true? Si

DL = b + y then, after reduc

FE = HK = x =( 3 ) - 2ax 4- x^

KL = EM = y or (4 ) 2ax - x^ =

EK = FL = h shows that the ai

(a^ = sq. DP) is
area by which b^ is increased. See f
increase always equals the decrease.

But a^ - 2x(a - x)

= (a - x)® approaches

0 when x approaches a in value.

(5) (a - x)^ = 0, when x = a, which is true
and (6) b^ 4* 2by 4* y^ = (b 4- y)^ = c^, when x = a,
for when x becomes a, (b 4- y) becomes c, and so, we

88

THE PYTHAGOREAN PROPOSITION

have which is true.

equation (2) Is true; it rests on the eq’s
(5) and (6), both of which are true.

.% whether a < = or > b, h^ = a^ + b^.
a. Devised by the author, in Dec. 1925. Also
a like proof to the above Is that of A. R. Colburn,
devised Oct. 18, 1922, and Is No. 96 In his collec-
tion of 108 proofs.

F, — Al iebralc-Geometric Proofs

In determining the equivalency of areas these
proofs are algebraic; but In the final comparison of
areas they are geometric.

One^Hundred

The construction, see
f'ig. 95 ^ being made, we have
sq. FE = (a + b}^.

But sq. FE = sq. AC
+ 4 trl. ABH

= + 4 -^ = h® + Sab.

Equating, we have

h® ■(- Sab = (a + b)® = a® + Sab
+ b®. h® = a® + b®,

a. See Scl. Am. Sup.,
V. 70, p. 582, Dec. 10, 1910,
credited to A. R. Colburn,
Washington, D.C.

SiQ&.!iyi!ldred.(^ne

Let AD = AG = X, HG = HC = y, and BC = BE
= z. Then AH = x + y, and BH = y + z.

With A as center and AH as radius describe

arc HE; with B as center and BH as radius describe

arc HD; with B as center, BE as radius describe arc

EC; with A as center, radius AD, describe arc DG.

Fig. 95

NICHOLAS COPERNICUS
1473-1543

ALGEBRAIC PROOFS

89

^ "f a'

> * t* I

i>y- »

Draw the parallel
lines as indicated. By in-
specting the figure it be-
comes evident that if y^= 2xz,
then the theorem holds. Now,
since AH is a tangent and AR
is a chord of same circle,

^2 _ «T> „ »-n __ (x + y)®

+ 2xy + 2xz.

AH^ = AR X ad,
= x(2y + 2z) =

or

^2

Whence y = 2xz.

sq. AK = [(x® + y® + 2xy)

= aq. AL] + [ (z® + 2yz +

Pig. 96 (2xz = y®)] = sq. HP. h®

= a + b .

a. See Sci. Am. Supt., V. 84, p. 3^2, Dec. 8,
1917 j and credited to A. R. Colburn. It is No. 79
in his (then) 91 proofs.

b. This proof is a fine Illustration of the
flexibility of geometry. Its value lies, not in a
repeated proof of the many times established fact,
but in the effective marshaling and use of the ele-
ments of a proof, and even more also in the better
insight which it gives us to the Interdependence of
the various theorems of geometry.

d D. e -H y. tl S C.® d -Iw ft

Draw the bisectors of
angles A, B and H, and from
their common point C draw the
perp*s OR, CX and CT; take AN
= AU = AP, and BZ = BP, and
draw lines UV par. to AH, NM
par. to AB and SY par. to BH.
Let AJ = AP = X, BZ = BP = y,
and HZ = HJ = z = CJ = CP
= CZ.

Now 2 tri. ABH = HB
X HA = (x + z ) (y + z ) = xy
+ xz + yz + = rect. PM

+ rect. HW + rect. HQ + sq. SX.

90

THE PYTHAGOREAN PROPOSITION

But 2 tri. ABH = 2AP x cP + 2BP x CP + (2 sq.
HO = 2PC®) = 2xz + 2yz + 2z®

= 2rect. HW + 2 rect. HQ + 2 sq. SX.

/. rect. PM = rect. HW + rect. HQ + sq. KX.

Now sq. AK = (sq. AO = sq. AW) + (sq. OK
= sq. BQ) + (2 rect. PM = rect. HW + 2 rect. HQ
+ 2 sq. SX) = sq. HG + sq. HD. h® = a® + b^.

a. This proof was produced by Mr. P. S. Smed-
ley, a photographer, of Berea, 0., June 10, 1901.

Also see Jury Wlpper, l880, p. fig* 31^
credited to E. Mbllmann, as given In "Archives d.
Mathematlk, u. Ph. Grunert," I851, for fundamentally
the same proof.

Let HR = HE = a = SG.
Then rect. GT = rect. EP,
and rect. RA = rect. QB.

.% trl*s 2, ^ and

5 are all equal. sq. AK

= h^ = (area of k trl. ABH
+ area sq. OM) = 2ba
+ (b - a )^ = 2ab + b^ - 2ba

+ a^ = b^ + a^ . h^ = a^

+ b^. Q.E.D.

a. See Math. Mo. ,

1858-9, Vol. I, p. 561,

where above proof Is given
by Dr. Hutton (tracts, Lon-
don, 1812, 5 vol*s, 820) In
his History of Algebra.

Take AN and AQ = AH, KM and KR = BH, and
through P and Q draw PM and QL parallel to AB; also
draw OR and NS par. to AC. Then CR = h - a, SK = h
- b and RS = a + b - h.

ALCTBRAIC PROOFS

91

jTJL

l_ _ I L. J.,

G K S

Fig. 99

Now aq. AK = CK® = 03® + RK®

- RS® + 2CR X SK, or h® = b® + a®

- (a + b - h)® + 2(h-a) x (h-b)

= b® + a® - a® - b® - h® - 2ab + 2ah
+ 2bh + 2h® - ah - 2bh + 2ab. 2CR
X SK = RS®, or 2(h - a)(h - b)

= (a + b - h)®, or 2h® + 2ab - 2ah

- 2bh = a® + b® + h® + 2ab + 2ah

- 2bh. h® = a® + b®,

a. Original with the author,
April 23, 1926.

G. — Algebraic-Geometric Proofs Throuih Similar Poly-
ions Other Than Squares,

1st, — Similar Trianiles

fi!l®.-Hundred_F£ve

Trl*s ACB, BDH and
HEA are three similar trl's
constructed upon AB, BH and
HA, and AK, BM and HO are
three corresponding rect*s,
double In area to trl's ACB,
BDH and HEA respectively.

Trl. ACB : trl. BDH
: trl. HEA = h® : a® : b®

= 2 trl. ACB ; 2 trl. BDA
= 2 trl. HEA = rect. AK
Produce LM and ON to their In-
tersection P, and draw PHG. It Is perp. to AB, and
by the Theorem of Pappus, see fig. l43, PH = QG.
by said theorem, rect. BM + rect. HO = rect. AK.
trl. BDH + trl. HEA = trl. ACB. h® = a® + b®,
a. Devised by the author Dec. 7, 1933.

(la£.!iiL!id!:£d_six

In fig. 100 extend KB to R, intersecting LM
at S, and draw PR and HT par. to AB. Then rect. BLMH
= paral. BSPH = 2 trl. BPH = 2 trl (BPH - PH x QB)

= rect. QK. In like manner, 2 trl. HEA - rect. AO.

: rect. BM : rect. HO.

ALGEBRAIC PROOFS

93

As tri*s AOB, BPH and HRA are similar isosce-
les tri * s , it follows that these tri*s are a particu-
lar case of proof One Hundred Six ,

And as tri. ABH : tri. BHQ : tri. HAQ = h^

: a® ; b® = tri. AOB : tri. BPH ; tri, HRA = penta-
gon 0 : pentagon P : pentagon R, since tri. ABH
= tri. BHQ + tri. HAQ. polygon 0 = polygon P
+ polygon R. h^ = a^ + b^.

a. Devised by the author Dec. 7> 1933.

fid® - H U. Q.d C® d -tli He

Upon the three sides of
the rt. tri. ABH are constructed
the three similar polygons (hav-
ing five or more sides — five in
fig. 103), ACDEB, BPGKH and HLMNA.
Prove algebraically that h^ = a^

+ b^, through proving that the
sum of the areas of the two less-
er polygons = the area of the
greater polygon.

In general, an algebraic
proof is impossible before trans-
formation. But granting that h^

= a^ + b^, it is easy to prove
that polygon (l ) + polygon (2 ) = polygon (3), as we
know that polygon (l ) ; polygon (2 ) ; polygon (3 )

= a^ : b^ : h^. But from this it does not follow
that a^ + b^ = h^.

See Beman and Smith* s New Plane and Solid
Geometry (l899), p. 211, exercise 438.

But an algebraic proof is always possible by
transforming the three similar polygons into equiva-
lent similar paral*s and then proceed as in proof
One Hundred Six .

Knowing that tri. ABH : tri. BHQ : tri. HAQ
= h® : a® : b®. — (l)

and that P. (?) : P. (l) : P. (2). [P = polygon]

» h® : a® : b®. (2); by equating tri. ABH : tri.

BHQ : tri, HAQ = P. (?) ; P. (l) : P. (2). But

94 THE PYTHAGOREAN PROPOSITION

tri. ABH = tri. BHQ + trl. HAQ. P. (5) = P. (l )

+ P. (2). h® = + b®. Q.E.D.

a. Devised by the author Deo. ^ , 1955.

b. Many more algebraic proofs are possible.

To evolve an original
demonstration and put it in a
form free from criticism is
not the work of a tyro.

3

II. GEOMETRIC PROOFS

All geometric demonstrations must result from
the comparison of areas — the foundation of which Is
superposition.

As the possible number of algebraic proofs
has been shown to be limitless, so It will be conclu-
sively shown that the possible number of geometric
proofs through dissection and comparison of congru-
ent or equivalent areas Is also "absolutely unlimit-
ed."

The geometric proofs are classified under
ten type forms, as determined by the figure, and only
a limited number, from the Indefinite many, will be
given; but among those given will be found all here-
tofore (to date, June 19^0), recorded proofs which
have come to me, together with all recently devised
or new proofs.

The references to the authors In which the
proof, or 'figure. Is found or suggested, are arranged
chronologically so far as possible.

The Idea of throwing the suggested proof Into
the form of a single equation Is my own; by means of
It every essential element of the proof Is set forth,
as well as the comparison of the equivalent or equal
areas.

The wording of the theorem for the geometric
proof Is: The square described upon the hypotenuse
of a riiht-aniled triangle is equal to the sum of the
squares described upon the other two sides.

TYPES

It Is obvious that the three squares con-
structed upon the three sides of a right-angled tri-
angle can have eight different positions, as per se-
lections. Let us designate the square upon the

97

98

THE PYTHAGOREAN PROPOSITION

hypotenuse by h, the square upon the shorter side by
a, and the square upon the other side by 5, and set
forth the eight arrangements; they are:

A. All squares h, a and b exterior.

B. a and b exterior and h interior.

G. h and a exterior and b interior.

D. h and b exterior and a interior.

E. a exterior and h and b interior.

P. b exterior and h and a interior.

G, h exterior and a and b interior.

H, All squares h, a and b interior.

The arrangement designated above constitute
the first eight of the following ten geometric types,
the other two being:

I, A translation of one or more squares.

J, One or more squares omitted.

Also for some selected figures for proving
Euclid I, Proposition 47, the reader is referred to
H, d'Andre, N. H. Math, (l846), Vol, 5, P. 524.

Note . By "exterior" is meant constructed
outwardly.

By interior" is meant constructed
overlapping the given right triangle.

A

This type includes all proofs derived from
the figure determined by constructing squares upon
each side of a right-angled triangle, each square be-
ing constructed outwardly from the given triangle.

The proofs under this type are classified as

follows :

(a) Those proofs in which pairs of the dis-
sected parts are congruent.

Congruency implies superposition, the most
fundamental and self-evident truth found in plane
geometry.

GEOMETRIC PROOFS

99

As the vays of dissection are so various, it
follows that the number of "dissection proofs" is un-
limited.

(b) Those proofs in which pairs of the dis-
sected parts are shown to be equivalent.

As geometricians at large are not in agree-
ment as to the symbols denoting "congruency" and
"equivalency" (personally the author prefers = for
congruency, and = for equivalency), the symbol used
herein shall be =, the context deciding its import.

(a) PROOFS IN WHICH PAIRS OF THE DISSECTED PARTS ARE
CONGRUENT.

Paper Folding "Proofs," Only Illustrative

Cut out a square piece of
paper EF, and on its edge, using
the edge of a second small square
of paper, EH, as a measure, mark
off EB, ED, LK, LG, FC and QA.

Fold on DA, BG, KN, KC,
CA, AB and BK. Open the sq. EF
and observe three sq's, EH, HF
and BC, and that sq. EH = sq. KG.

With scissors cut off
tri. CFA from sq. HF, and lay it
on sq. BC in position BHA, ob-
serving that it covers tri. BHA of sq. BC; next cut
off KLC from sq*s NL and HF and lay it on sq. BC in
position of KNB so that MG falls on PO. Now observe

that tri. KMN is part of sq. KG and sq. BC and that

the part HMCA is part of sq. HF and sq. BC, and that

all of sq. BC is now covered by the two parts of sq.

KG and the two parts of sq. HF.

Therefore the (sq. EH = sq. KG) + sq. HF
= the sq. BC. Therefore the sq. upon the side BA
which is sq, BC = the sq, upon the side BH which is

100

THE PYTHAGOREAN PROPOSITION

sq. BD + the sq. upon the side HA which Is sq, HF.

/. h^ = + h^, as shown with paper and scissors,

and ohservatlon .

a. See "Geometric Exercises in Paper Fold-
ing," (T. Sundra Row*s), I905, p. l4, fig. 15, by
Beman and Smith; also School Visitor, 1882, Vol. Ill,
p. 209; also F. C. Boon, B.H., in "A Companion to
Elementary School Mathematics," (1924), p. 102,
proof 1.

Iwo

Fig. 105

exactly cover sq. BC.
= EL) + sq. upon AH.

Cut out three sq*s
EL whose edge is HB, PA
whose edge HA, and BC whose
edge is AB, making AH=2HB.

Then fold sq. PA
along MN and OP, and sepa-
rate into 4 sq's MP, QA, ON
and PQ each equal to sq. EL.

Next fold the 4 pa-
per sq's (U, R, S and T be-
ing middle pt's), along HU,
PR, QS and MT, and cut,
forming parts, 1, 2, 5, 4,
5> 6, 7 and 8.

Now place the 8
parts on sq. BC in posi-
tions as indicated, reserv-
ing sq. 9 for last place.
Observe that sq. PA and EL

sq. upon BA = sq. upon (HB
h® = a® + b®. Q.E.P.

a. Beman and Smith's Row's (1905), work,
p. 15» fig. 14 j also School Visitor, 1882, Vol. Ill,
p. 208; also P. C. Boon, p, 102, proof 1.

GEOMETRIC PROOFS

101

Three

Cut out three sq’s as in
fig. 105. Fold small sq. 9 (fig*
105) along middle and cut, form-
ing 2 rect’s; cut each rect.
along diagonal, forming 4 rt.
trl*s, 1, 2, 5 and 4. But from
each corner of sq. FA (fig. 105),
a rt . tri. each having a base HL
= ^HP (fig. 105; FT = iFM), giv-
ing 4 rt. trl^s 5 , 6, 7 and 8
(fig. 106), and a center part 9
(fig. 106), and arrange the pieces as in fig. IO6,
and observe that sq. HC = sq. EL + sq. HG, as in fig.
105. h^ = a® + b^.

a. See "School Visitor,” 1882, Vol. Ill,

p. 208.

b. Proofs Tvo and Three are particular and
illustrative — not general — but useful as a paper and
scissors exercise.

c. With paper and scissors, many other proofs,
true under all conditions, may be produced, using
figs. 110, 111, etc., as models of procedure.

Eour

Particular case--ll-
lustratlve rather than demon-
strative.

The sides are to each
other as 3 f 4, 5 units. Then
sq. AK contains 25 sq. units,
HD 9 sq. units and HG I6 sq.
units. Now it is evident that
the no. of unit squares in the
sq. AK = the sum of the unit
squares in the squares HD and
HG.

/. square AK ss sq. HD + sq. HG.

Fig. 107

102

THE PYTHAGOREAN PROPOSITION

a. That by the use of the lengths 4, and
5, or length having the ratio of 5 : 4 ; 5^ a right-
angled triangle Is formed was known to the Egyptians
as early as 2000 B.C., for at that time there existed
professional "rope-fasteners”; they were employed to
construct right angles which they did by placing
three pegs so that a rope measuring off 3 , 4 and 5
units would just reach aro\md them. This method is
in use today by carpenters and masons; sticks 6 and

8 feet long form the two sides and a "ten-foot" stick
forms the hypotenuse, thus completing a right-angled
triangle, hence establishing the right angle.

But granting that the early Egyptians formed
right angles in the "rule of thumb" manner described
above, it does not follow, in fact it is not be-
lieved, that they knew the area of the square upon
the hypotenuse to be equal to the sirni of the areas of
the squares upon the other two sides.

The discovery of this fact is credited to
Pythagoras, a renowned philosopher and teacher, born
at Samos about 570 B.C., after whom the theorem is
called "The Pythagorean Theorem." (See p. 5).

b. See Hill*s Geometry for Beginners, p. 155;
Ball*s History of Mathematics, pp. 7-10; Heath* s
Math. Monographs, No. 1, pp. 15-17; The School Visi-
tor, Vol. 20, p. 167.

EiY^

Another particular
case is illustrated by fig.
108, in which BH = HA, show-
ing 16 equal triangles.

Since the sq. AK con-
tains 8 of these triangles,

/. sq. AK = sq. HD + sq. HG.

/. h^ = a^ + b^.

a. For this and many
other demonstrations by dis-
section, see H. Perigal, in
Messenger of Mathematics,

GEOMETRIC PROOFS

105

1873, V. 2, p. 105; also see Pourrey, p. 68.

b. See Beman and Smith* s New Plane and Solid
Geometry, p. IO5, fig. 1.

c. Also R. A. Bell, Cleveland, 0., using sq.
AK and lines AK and BC only.

Six

In fig. 108, omit lines AP, BE, LM and NO,
and draw line PE; this gives the fig. used In "Grand
Lodge Bulletin," Grand Lodge of Iowa, A.P. and A.M.,
Vol. 50, Peb. 1929^ No. 2, p. 42. The proof Is ob-
vious, for the 4 equal Isosceles rt. trl*s which
make up sq. PB = sq. AK. /. h^ = a^ + b^.

a. This gives another form for a folding pa-
per proof.

S^yen

In fig. 108, omit lines as In proof Six , and
It Is obvious that trl*s 1, 2, 5 and 4, In sq*s HG
and HD will cover trl*s 1, 2, 5 and 4 In sq. AK, or
sq. AK = sq. HD + sq. HG. h^ = a^ + b^.

a. See Versluys (1914), fig. 1, p. 9 of his
96 proofs.

Elflht

In fig. 109, let
HAGP denote the larger
sq. HG. Cut the smaller
sq. EL Into two equal
rectangles AN and ME, fig
109, and form with these
and the larger sq. the
rect. HDEP. Produce DH
so that HR = HP. On RD
as a diameter describe a
semicircle DCR. Produce

104

THE PYTHAGOREAN PROPOSITION

HP to C In the arc. Join CD, cutting PG In P, and
AG In S. Complete the sq. HK.

Now trl*s CPP and LED are congruent as are
trl*s CKL and PED. Hence sq. KH = (sq. EL, fig. 105
= rect. AN + rect. ME, fig. IO9) + (sq. HG, fig. 105
= quad. HASPP + trl. SGP, fig. 109). /. h^ = a^ + h^.

a. See School Visitor, I882, Vol. Ill, p. 208.

b. This method, embodied in proof Eight , will
transform any rect. Into a square.

c. Proofs Two to Eight Inclusive are Illus-
trative rather than demonstrative.

Demonstrative Proofs

Nine

In fig. 110, through
P, Q, R and S, the centers of
the sides of the sq. AK draw
^ PT and RV par. to AH, and QU
V and SW par. to BH, and through
0, the center of the sq. HG,
draw XH par. to AB and lY
par. to AC, forming 8 congru-
ent quadrilaterals; viz., 1,

2, 5 and 4 In sq. AK, and 1,

2, 5 and 4 In sq. HG, and sq.

5 in sq. AK = sq. (5 = HD).

The proof of their congruency
Is evident, since, in the
paral. OB, (SB = SA) = (OH
= OG = AP since AP = AS).

(Sq. AK = 4 quad. APTS + sq. TV) = (sq. HG = 4 quad.

OYHZ) + sq. HD. /. sq. on AB = sq. on BH + sq. on AH.

/. h^ = a^ + b^.

a. See Mess. Math., Vol. 2, 1873, p. 104, by
Henry Perlgal, P. R. A. S., etc., Macmillan and Co.,
London and Cambridge. Here H. Perlgal shows the
great value of proof by dissection, and suggests Its
application to other theorems also. Also see Jury

GEOMETRIC PROOFS

105

Wlpper, 1880, p. 50, fig. 46; Ebene Geometrle, Von G.
Mahler, Leipzig, 1897, p. 58, fig. 7I, and School
Visitor, V. Ill, 1882, p. 208, fig. 1, for a particu-
lar application of the above demonstration; Versluys,
1914, p. 37, fig. 37 taken from "Plane Geometry" of
J. S. Mackay, as given by H. Perlgal, I83O; Pourrey,
p. 86; F. C. Boon, proof 7, p. 105; Dr. Leltzmann,
p. 14, fig. 16.

b. See Todhunter*s Euclid for a simple proof
extracted from a paper by De Morgan, in Vol. I of the
Quarterly Journal of Math., and reference is also
made there to the work "Der Pythagoraische Lehrsatz,"
Mainz, 1821, by J. J. I. Hoffmann.

c. By the above dissection any two squares
may be transformed into one square, a fine puzzle for
pupils in plane geometry.

d. Hence any case in which the three squares
are exhibited, as set forth under the first 9 types
of II, Geometric Proofs, A to J inclusive (see Table
of Contents for said types) may be proved by this
method.

c. Proof Nine is unique in that the smaller
sq. HD is not dissected.

Ikn

In fig. Ill, on CK
construct tri. CKL = trl. ABH;
produce CL to P making LP = BH
and take LN = BH; draw NM, AO
and BP each perp. to CP; at
any angle of the sq. QH, as F,
construct a tri. GSP = tri.
ABH, and from any angle of the
sq. HD, as H, with a radius
= KM, determine the pt. R and
draw HR, thus dissecting the
sq*s, as per figure.

It is readily shown

Fig. Ill

106

THE PYTHAGOREAN PROPOSITION

that sq. AK = (trl. CMN = trl. BTP ) + (trap. NMKL
= trap. DRHB) + (trl. KTL = trl. HRE) + (quad. AOTB
+ trl. BTP = trap. GAHS) + (trl. AGO = trl. GSP)

= (trap. DRHB + trl. HRE = sq. BE) + (trap. GAHS
+ trl. GSP = sq. AP) = sq. BE + sq, AP. /. sq. upon
AB = sq. upon BH + sq. upon AH. /. = a^ + b^.

a. This dissection and proof were devised by
the author, on March l8, 1926, to establish a Law of
Dissection, by which, no matter how the three squares
are arranged, or placed, their resolution Into the
respective parts as numbered In fig. Ill, can be read-
ily obtained.

b. In many of the geometric proofs herein the
reader will observe that the above dissection, wholly
or partially, has been employed. Hence these proofs
are but variation of this general proof.

+ (sq. LN = sq. RH)
+ sq. RH = sq. BE +
BH + sq. upon AH. .

a. Original

sq

In fig. 112, con-
ceive rect. TS cut off from
sq, AP and placed in posi-
tion of rect. OP, AS co-
inciding with HE; then DEP
Is a st. line since these
rect. were equal by construc-
tion. The rest of the con-
struction and dissection Is
evident .

sq. AK = (trl. CKN = trl.

PBD) + (trl. KBO = trl. BPQ)

+ (trl. BAL = trl. TPQ)

+ (trl. ACM = trl. FTG)
sq, BE + rect, QE + rect. GQ
GH. /. sq, upon AB = sq. upon
+ b^.

. h^ = a-=

with the author after having care-
fully analyzed the esoteric implications of Bhaskara's
"Behold!” proof— see proof Two Hundred Twenty-Pour,
fig. 325.

GEOMETRIC PROOFS

107

b. The reader vlll notice that this dissec-
tion contains some of the elements of the preceding
dissection, that it is applicable to all three-square
figures like the preceding, but that it is not so
simple or fundamental, as it requires a transposition
of one part of the sq. GH, — the rect. TS--, to the
sq. HD, --the rect. in position QE--, so as to form
the two congruent rect*s GQ and QP,

c. The student will note that all geometric
proofs hereafter, which make use of dissection and
congruency, are fundamentally only variations of the
proofs established by proofs Nine , Ten and Eleven and
that all other geometric proofs are based, either par-
tially or wholly on the equivalency of the correspond-
ing pairs of parts of the figures under consideration.

Twelve

This proof is a sim-
ple variation of the proof
Ten above. In fig. 115# ex-
tend GA to M, draw ON and BO
perp, to AM, take NP = BD
and draw PS par. to CN, and
through H draw QR par. to AB.
Then since it is easily
shown that parts 1 and 4 of
sq. AK = parts 1 and 4 of
sq. HD, and parts 2 and 3 of
sq. AK = 2 and 3 of sq. HG,

/. sq. upon AB = sq. upon BH
+ sq. upon AH.

a. Original with

the author March 28, 1926 to obtain a figure more
readily constructed than fig. 111.

b. See School Visitor, l882, Vol. Ill, p.
208-9; Dr. Leltzmann, p. 15# fig. 17# ^th Ed*n.

108

THE PYTHAGOREAN PROPOSITION

Ihirteen

In fig. Il4, produce
CA to 0, KB to M, GA to V,
making AV = AG, DB to U, and
draw KX and CW par. resp. to
BH and AH, GN and HL par. to
AB, and OT par. to PB.

Sq. AK = [tri. CKW = trl.

(HLA = trap, BDEM + trl. NST)]
+ [tri. KBX = tri. GNF
= (trap. OQNP + trl. BMH)]

+ (trl. BAU = trl. OAT)

+ (tri. ACV = tri. AOG)

+ (sq. VX = paral. SN)

= sq. BE + sq. HG. sq.

upon AB = sq. upon BH + sq. upon AH.

a. Original with author March

h'

28,

= a

1926,

9:50

p .m.

b. A variation of the proof Eleven above.

Fourteen

Fig. 115

+ sq. ST = sq. BE + sq.
BH + sq. upon AH. .’. h^

Produce CA to S,
draw SP par. to PB, take HT
= HB, draw TR par. to HA,
produce GA to M, making AM
= AG, produce DB to L, draw
KO and ON par, resp. to BH
and AH, and draw QD. Rect.
RH = rect. QB. Sq. AK
= (trl. CKN = trl. ASG)

+ (tri. KBO = trl. SAQ)

+ (trl. BAL = trl. DQP )

+ (trl. ACM = tri. QPE)

4 - (sq. LN = sq. ST) = rect.
PE + rect. GQ + sq. ST = sq.
BE + rect. QB + rect. GQ
GH. sq. upon AB = sq. upon
= a^ + b^.

GEOMETRIC PROOFS

109

a. Original with author March 28, 1926, 10

a.m.

b. This is another variation of fig. 112.

Fifteen

Take HR = HE and

PS = PR = EQ = DP.

Draw RU par. to
AH, ST par. to PH, QP par.
to BH, and UP par. to AB.
Extend GA to M, making AM
= AG, and DB to L and draw
CN par. to AH and KO par.
to BH.

Place rect. GT in
position of EP. Obvious
that: Sq. AK = parts (l

+2+3)+ (4+5 of rect.
HP). /. Sq. upon AB = sq.

Fig. Il6 upon BH + sq. upon AH.

h^ = a^ + b^.

a. Math. Mo., 1858-9, Vol. I, p. 231, where
this dissection is credited to David W. Hoyt, Prof.
Math, and Mechanics, Polytechnic College, Phila., Pa.;
also to Pliny Earle Chase, Phila., Pa.

b. The Math. Mo. was edited by J. D. Runkle,
A.M., Cambridge Eng. He says this demonstration is
essentially the same as the Indian demonstration
found in ”Blja Gaulta" and referred to as the figure
of "The Brides Chair."

c. Also see said Math. Mo., p. 361, for an-
other proof; and Dr. Hutton (tracts, London, l8l2, in
his History of Algebra).

^ixtgen

In fig. 117, the dissection is evident and
shows that parts 1, 2 and 3 in sq. AK are congruent
to parts 1, 2 and 3 in sq. HG; also that parts 4 and

110

THE PYTHAGOREAN PROPOSITION

5 in sq. AK are congruent to
parts 4 and 5 in sq. HD.

/. (sq. AK = parts 1 + 2 + 5
+ 4 + 5) = (sq. HG = parts
1 + 2 + 5) + (sq* HD = parts
4 + 5). /. sq. on AB = sq.

on AH. /.

on BH

■» 2 I

sq.

.2

= a^ + b"

a. See Jury Wlpper,
1880, p. 27, fig. 24, as
given by Dr. Rudolf Wolf in
"Handbook der Mathematik,
etc.," 1869; Journal of Ed-
ucation, V. XXVIII, 1888,

p. 17, 27th proof, by C. W. Tyron, Louisville, Ky.;
Beman and Smithes Plane and Solid Geom. , 1895^ P*
fig. 5; Am. Math. Mo., V. IV, l897> P* 1^9 proof
XXX7X; and Heath* s Math. Monographs, No. 2, p. 55,
proof XXII. Also The School Visitor, V. Ill, 1882,
p. 209, for an application of it to a particular case;
Pourrey, p. 87, by Ozanam, 1778, R. Wolf, 1869.

b. See also "Recreations in Math, and Phys-
ics," by Ozanam; "Curiosities of Geometry," 1778, by
Zle E. Pourrey; M. Krdger, I896; Versluys, p. 59,
fig. 59, and p. 4l, fig. 4l, and a variation is that
of Versluys (l9l4), p. 40, fig. 4l.

Seventeen

Extend CA to M and
KB to Q, draw MN par. to AB.
Extend GA to T and DB to 0.
Draw CP par. to AB. Take
OR = HB and draw RS par. to
HB.

Obvious that sq. AK
= sum of parts (4+5)

+ (1 + 2 + 5) = sq. HD + sq.
HG. sq. upon AB = sq.
upon BH + sq. upon HA.

= a^ + b®. Q.E.D.

GEOMETRIC PROOFS

111

a. Conceived by the author, at Nashville, 0.,

March 26, for a high school girl there, while

present for the funeral of his cousin; also see
School Visitor, Vol. 20, p. I67.

b. Proof and fig. II8, Is practically the
same as proof Sixteen , fig. 117-

On Dec. 17, 1939, there came to me this: Der
Pythagorelsche Lehr sat s von Dr. W. Leltzmann, 4th
Edition, of 1930 (1st Ed*n, I91I, 2nd Ed*n, 1917,

3rd Ed*n, ), In which appears no less than 23

proofs of the Pythagorean Proposition, of which 21
were among my proof herein.

This little book of 72 pages Is an excellent
treatise, and the bibliography, pages 70, 71, 72, Is
valuable for Investigators, listing 21 works re this
theorem.

My manuscript, for 2nd edition, credits this
work for all 23 proof therein, and gives, as new
proof, the two not Included In the said 21.

Eighteen

credited to any one, but
proofs.

In fig. 119, the

dissection Is evident and
shows that parts 1, 2 and 3
In sq. HG are congruent to
parts 1, 2 and 3 In rect.
QC; also that parts 4, 5, 6
and 7 In sq. HD are congru-
ent to parts 4, 5, 6 and 7
In rect. QR.

Therefore, sq. upon
AB = sq. upon HB + sq. upon
HA. h® = a® + b®. Q.E.D.

a. See dissection,
Tafel II, In Dr. W. Leltz-
mann* s work, 1930 ed*n — on
last leaf of said work. Not
Is based on H. Dobrlner*s

112

THE PYTHAGOREAN PROPOSITION

nineteen

In fig. 120 draw GD,
and from F and E draw lines to
GD par. to AC; then extend DB
and GA, forming the rect. AB;
through C and K draw lines par.
respectively to AH and BH, form-
ing tri’s equal to tri. ABH.
Through points L and M draw
line par. to GD. Take KP = BD,
and draw MP, and through L
draw a line par. to MP.

Number the parts as in
the figure. It is obvious that
Fig. 120 the dissected sq’s HG and HD,

giving 8 triangles, can be ar-
ranged in sq. AK as numbered; that is, the 8 tri*s
in sq. AK can be superimposed by their 8 equivalent
trl*s in sq*s HG and HD. sq. AK = sq. HD + sq. HG.
h® = + b®. Q.E.D.

a. See dissection, Tafel I, in Dr. W. Leltz-
mann work, 1930 ed*n, on 2nd last leaf. Not credited
to any one, but is based on J. E. Bbttcher^s work.

Twenty

In fig. 121 the con-
struction is readily seen,
as also the congruency of
the corresponding dissected
parts, from which sq. AK
= (quad. CPNA = quad. LAHT)
+ (tri. CKP = tri. ALG)

+ (tri. BOK = quad. DEHR
+ tri. TPL) + (tri. NOB
= tri. RBD).

sq. upon AB = sq.
upon BH + sq. upon AH.

Fig. 121

GEOMETRIC PROOFS

113

a. See Math. Mo., V. IV, l897> p. l69> proof

XXXVIII.

Iwentif-ine

The construction
and dissection of fig. 122
Is obvious and the congru-
ency of the corresponding
parts being established, and
we find that sq. AK = (quad.
ANMR = quad. AHWX) + (trl.
CNA = trl. WPG) 4 - (trl. CQM
= trl. AXG) + (trl. MQK
= trl. EDU) + (trl. POK
= trl . THS ) + (pentagon BLMOP
= pentagon ETSBV ) + (trl.

BRL = trl . DUV ) . sq . upon

AB = sq. upon BH + sq. upon
AH. = a® + b®.

a. Original with the author of this work,
August 9s 1900. Afterwards, on July 4, 1901, I found
same proof In Jury Wlpper, l880, p. 28, fig. 25 > as
given by E. von Llttrow In "Popularen Geometrle,”
1859; also see Versluys, p. 42, fig. 45.

Iwen tjf-Iwfi

Extend CA to Q, KB
to P, draw RJ through H, par.
to AB, HS perp. to OK, SU
and ZM par, to BH, SL and ZT
par. to AH and take SV = BP,
DN = PE, and draw VW par. to
AH and NO par. to BP.

Sq. AK = parts (l + 2
+ 5 + 4 = sq. HD) + parts
(5+6+7 = sq. HG); so dis-
sected parts of sq, HD + dis-
sected parts of sq. HG (by
superposition), equals the
dissected parts of sq. AK.

114

THE PYTHAGOREAN PROPOSITION

Sq. upon AB = sq. upon BH + sq. upon AH*

+ b®. Q.E.D.

a. See Veraluys, p. 45, fig. 44.

b. Pig. and proof, of Tventy-Tvo is very much
like that of Twenty- One .

Twenty-Three

Fig. 124

After showing that
each numbered part foxmd In
the sq*s HD and HG Is congru-
ent to the corresponding num-
bered part In sq. AK, which
Is not difficult. It follows
that the sum of the parts In
sq. AK = the sum of the parts
of the sq. HD + the Siam of
the parts of the sq. HG.

the sq. upon AK
= the sq. upon HD -f- the sq.
upon HA. h^ = a^ + b^.
Q.E.D.

a. See Geom. of Dr.

H. Dobrlner, I 898 ; also Versluys, p. 45> fig. 46,
from Chr. Nielson; also Leltzmann, p. 15 , fig. 15 ,
4th Ed*n.

Proceed as In fig.
124 and after congruency Is
established. It Is evident
that, since the eight dis-
sected parts of sq. AK are
congruent to the correspond-
ing numbered parts found In
sq*s HD and HG, parts (l + 2
+3+4+5+6+7+8ln
sq. AK) = parts (5+6+7
+ 8) + (1 + 2 + 3 + 4) In
sq» s HB and HG.

GEOMETRIC PROOFS

115

aq. upon AB = sq. upon HD + sq. upon HA,

/. + b®.

a. See Paul Epstein’s (of Straatsberg), col-
lection of proofs; also Versluys, p. 44, fig. 45;
also Dr. Leltzmann’s 4th ed’n, p. 15, fig. l4.

Iwenty^-F £ve

Establish congruency
of corresponding parts; then
It follows that: sq. AK
(= parts 1 and 2 of sq. HD
+ parts 5> 4 and 5 of sq. HG)
= sq. HD + sq. HG. /. sq.
upon AB = sq. upon HD + sq.
upon HA. h^ = a^ + b^,
Q.E.D.

a. See Versluys, p.
58, fig. 58. This fig. Is
similar to fig. Ill,

Iwent^-Six

Since parts 1 and 2
of sq. HD are congruent to
like parts 1 and 2 in sq.
AK, and parts 5, 4, 5 and 6
of sq. HG to like parts 5,
4, 5 and 6 in sq. AK. /. sq.
upon AB = sq. upon HB + sq.
upon HA. /. h^ = a^ + b^.
Q.E.D

a. This dissection
by the author, March 26,
1935.

Fig. 127

116

THE PYTHAGOREAN PROPOSITION

Take AU and CV = BH
and drav UN par. to AB and
VT par. to BK; from T draw
TL par. to AH and TS par. to
BH, locating pts. L and S;
complete the sq*s LN and SQ,
making sides SR and LM par.
to AB. Draw SW par. to HB
and CJ par. to AH. The 10
parts found In sq*s HD and
HG are congruent to corre-
sponding parts In sq. AK.
the sq. upon AB = sq. upon HB
+ sq. upon HA. /. h^ = a^+
Q.E.D.

a. This proof, and dissection, was sent to
me by J . Adams, Chassestreet 51^ The Hague, Holland,
April 1933.

b. All lines are either perp. or par. to the
sides of the trl. ABH — a unique dissection.

c. It is a fine paper and scissors exercise.

— IK

Fig. 128

Iwenty-Eigh t

Fig. 129

Draw AP and BE; pro-
duce GA to P making AP = AG;
produce DB to 0; draw CQ par.
to AH and KR par. to BH; con-
struct sq. LN = sq. OQ; draw
PL and PN; take AT and KS
= to PM. Congruency of cor-
responding numbered parts hav-
ing been established, as is
easily done, it follows that:
sq. upon AB = sq. upon HB
+ sq. upon HA. h® = a^ + b®.
Q.E.D.

GEOMETRIC PROOFS

117

a. Benljr von Guthell, oberlehrer at Nurn-
berg, Germany, produced the above proof. He died In
the trenches in Prance, 191^. So wrote J. Adams
(see a, fig. 128), August 1933.

b. Let us call it the B. von Gutheil World
War Proof.

c. Also see Dr. Leitzmann, p. 15, fig. l8,
1930 ed»n.

Iw^nt^-Nine

In fig. 130, extend
CA to 0, and draw ON and KP
par. to AB and BH respective-
ly, and extend DB to R. Take
BM = AB and draw DM. Then
we have sq. AK = (trap. ACKP
= trap. OABN = pentagon
OGAHN) + (tri. BRK = trap.
BDLH + tri. MHL = tri. OPN)

+ (tri. PRB = tri. LED). /.
sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = a^+ b^.

a. See Math. Mo., V.
IV, 1897, P. 170, proof XLIV.

ItlLit

Pig. 151 objectifies
the lines to be drawn and
how they are drawn is readily
seen.

Since tri. OMN = tri.
ABH, tri. MPL = tri. BRH,
tri. BML = tri. AOG, and tri.
OSA = tri. KBS (K is the pt.
of intersection of the lines
MB and OS ) then sq . AK = trap.
ACKS + tri. KSB = tri. KOM
* trap. BMOS + tri. OSA
= quad. AHPO + tri. ABH

Pig. 131

118

THE PYTHAGOREAN PROPOSITION

+ trl. BML + trl. MPL = quad. AHPO + trl. OMN + trl.
AOG + trl, BRH = (pentagon AHPOG + trl. OPP) + (trap,
PMNP = trap. RBDE) + trl. BRH = sq. HG + sq. HD.
sq. upon AB = sq. upon HD + sq. upon AH. /. = a^+b^.

a. See Scl. Am. Sup., V. 70, p. 383, Dec. 10,
1910. It Is No. l4 of A. R. Colburn* 3 IO8 proofs.

Extend GA making AP
= AG; extend DB making BN
= BD = CP. Trl. CKP = trl.
ANB = -I sq. HD = -I rect. LK.
Trl. APB = ^ sq. HG = |- rect.
AM. Sq. AK = rect. AM
+ rect. LK.

sq, upon AB = sq.
upon HB + sq. upon AH.

= + b®. Q.E.D.

a. This Is Huygens*
proof (1657); see also Ver-
sluys, p. 25, fig. 22.

Fig. 133

Extend GA making AD
= AO. Extend DB to N, draw
CL and KM, Extend BP to S
making PS = HB, complete sq.
SU, draw HP par. to AB, PR
par. to AH and draw SQ.

Then, obvious, sq.

AK = 4 trl. BAN + sq. NL
= rect. AR + rect. TR + sq.
GQ = rect. AR + rect. QP
+ sq. GQ + (sq. TP = sq. ND)
= sq. HG + sq. HD. /. sq.
upon AB = sq. upon BH + sq.
upon AH. h® = a^ + b^.
Q.E.D.

EUCLID

Lived about 300 B*C

C3E0METRIC PROOFS

119

a. This proof Is credited to Miss E. A.
Coolldge, a blind girl. See Journal of Education,

V. XXV'III, 1888, p. 17, 26th proof.

b. The reader will note that this proof em-
ploys exactly the same dissection and arrangement as
found In the solution by the Hindu mathematician,
Bhaskara. See fig. 324, proof Two Hundred Twenty -
Five .

(b ) THOSE PROOFS IN WHICH PAIRS OP THE DISSECTED
PARTS ARE SHOWN TO BE EQUIVALENT.

As the triangle Is fundamental In the deter-
mination of the equivalency of two areas, Euclid's
proof will be given first place.

ItliLilrlllLi®

Draw HL perp. to CK,
and draw HC, HK, AD and BG.

Sq. AK = rect. AL + rect. BL

= 2 trl. HAC + 2 trl. HBK

= 2 trl. GAB + 2 trl. DBA

= sq. GH + sq. HD. sq. upon

AB = sq. upon BH + sq. upon AH.

a. Euclid, about 300
B.C. discovered the above
proof, and It has found a place
In every standard text on ge-
ometry. Logically no better
proof can be devised than Eu-
clid's.

For the old descrip-
tive form of this proof see Elements of Euclid by
Todhunter, 1887, Prop. 47, Book I. For a modern mod-
el proof, second to none, see Beman and Smith's New
Plane and Solid Geometry, 1899, p. 102, Prop. VIII,
Book II. Also see Heath's Math. Monographs, No. 1,
1900 , p. 18 , proof I; Versluys, p. 10, fig. 3, and
p. 76 , proof 66 (algebraic); Fourrey, p. 70 , fig. a;

120

THE PYTHAGOREAN PROPOSITION

also The New South Wales Freemason, Vol. XXXIII, No,
4, April 1, 1958, p. 178, for a fine proof of Wor.
Bro. W. England, P.S.P., of Auckland, New Zealand.
Also Dr. Leltzmann*s work (1930), p. 29, flg*s 29
and 30.

b. I have noticed lately two or three Ameri-
can texts on geometry In which the above proof does
not appear. I suppose the author wishes to show his
originality or Independence — possibly up-to-dateness.
He shows something else. The leaving out of Euclid's
proof Is like the play of Hamlet with Hamlet left
out.

c. About 870 there worked for a time. In Bag-

dad, Arabia, the celebrated physician, philosopher
and mathematician ibn fturra ibn Mervan (826-

901), Abu-Hasan, al-Harranl, a native Of Harran In
Mesopotamia. He revised ishaq ibn Honeiu»s transla-
tion of Euclid's Elements, as stated at foot of the
photostat .

See David Eugene Smith's "History of Mathe-
matlcs," (1923), Vol. I, pp. 171-3.

d. The figure of Euclid's proof. Fig, 1^4
above. Is known by the French as pon aslnorum, by the
Arabs as the "Figure of the Bride."

e. "The mathematical science of modern Europe

dates from the thirteenth century, and received Its
first stimulus from the Moorish Schools In Spain and
Africa, where the Arab works of Euclid, Archimedes,
Appollonlus and. Ptolemy were not uncommon "

"First, for the geometry. As early as 1120
an English monk, named Adelhard (of Bath), had ob-
tained a copy of a Moorish edition of the Elements of
Euclid; and another specimen was secured by Gerard of
Cremona In ll86. The first of these was translated
by Adelhard, and a copy of this fell Into the hands
of Giovanni Campano or Companus, who In 1260 repro-
duced It as his own. The first printed edition was
taken from It and was Issued by Ratdolt at Venice In
1482." A History of Mathematics at Cambridge, by
W. W. R. Ball, edition 1889, PP. 3 and 4.

geometric proofs

121

\ \

i\

i» i-^ ' w— ^.A' ^ ''J->)f >^‘ jt

-i-i'

J^LJi j'jiiij/i/ , ^ .

ki^ tiffj C^ V ^^*t> 2y J L

ly^, I 1 a-/

'‘^■f' wi^’^ U'l>

•' ^lip! uX>> rv iLj

,, u,y

• ^ ' »

PYTHAGOBEAN THEOREM IK TABIT IBN QORRA»S TRANSLATION
OF EUCLID

The translation vas made hy Is]^q Ihn 9oneln (died 910) but
was revised by TAblt Ibn Qorra, c. 89O. This manuscript was
written In I 55 O.

120a

THE PYTHAGOREAN PROPOSITION

Ihirt^-Fgur

Fig. 135

Extend HA to L making
AL = HE, and HB to N making
BN = HP, draw the perp. HM,
and join LC, HC, and KN. Ob-
viously trl*s ABH, CAL and
BKN are equal. /. sq. upon AK
= rect. AM + rect. BM = 2 trl.
HAG + 2 trl. HBK = HA x CL
-f- HB X KN = sq. HG + sq. HD.

sq. upon AB = sq, upon HB
J- sq. upon HA. = a^+ b^.

a. See Edwards* Geom.,
p. 155, fig. (4); Versluys,
p. 16, fig, 12, credited to
De Gelder (l 806 ).

b. ”To llliomlne and enlarge the field of con-
sciousness, and to extend the growing self. Is one
reason why we study geometry,”

"One of the chief services which mathematics
has rendered the human race in the past century Is
to put * common sense* where It belongs, on the top-
most shelf next to the dusty canister labeled * dis-
carded nonsense.*" Bertrand Russell.

c. "Pythagoras and his followers found the
ultimate explanation of things In their mathematical
relations. "

Of Pythagoras, as of Omar Khayyam:

"Myself when young did eagerly frequent
Doctor and Saint, and heard great argument
About It and about; but evermore
Came out by the same door where In I went."

GEOMETRIC PROOFS

121 a

HISTORY SAYS:

1. "Pythagoras, level-headed, wise man, went quite
mad over seven. He found seven sages, seven
wonders of the world, seven gates to Thebes, sev-
en heroes against Thebes, seven sleepers of
Ephesus, seven dwarfs beyond the mountains --and
so on up to seventy times seven."

2. "Pythagoras was Inspired- -a saint, prophet, found
er of a fanatically religious society."

3. "Pythagoras visited Ionia, Phonecia and Egypt,
studied In Babylon, taught in Greece, committed
nothing to writing and founded a philosophical
society. "

4. "Pythagoras declared the earth to be a sphere,
and had a movement in space . "

5. "Pythagoras was one of the nine saviors of clvlll
zatlon. "

6. "Pythagoras was one of the four protagonists of
modern science."

7. "After Pythagoras, because of the false dicta of
Plato and Aristotle, it took twenty centuries to
prove that this earth is neither fixed nor the
center of the universe."

8. "Pythagoras was something of a naturallst--he was
2500 years ahead of the thoughts of Darwin."

9. "Pythagoras was a believer in the Evolution of
man . "

10. "The teaching of Pythagoras opposed the teaching
of Ptolemy."

11. "The solar system as we know it today is the one
Pythagoras knew 2500 years ago."

12. "What touched Copernicus off? Pythagoras who
taught that the earth moved around the sun, a
great central ball of fire."

13 . "The cosmology of Pythagoras contradicts that of
the Book of Genesis — a barrier to free thought
and scientific progress."

14. "Pythagoras saw man- -not a cabbage, but an anl-
mal--a bundle of possibilities — a rational ani-
mal . "

15 . "The teaching of Pythagoras rests upon the Social
Ethical and Aesthetlcal Laws of Nature."

122

THE PYTHAGOREAN PROPOSITION

ItliLti-Fiye

Fig. 136

Draw HN par. to AC,

KL par. to BP, CN par. to AH,
and extend DB to M. It is
evident that sq. AK = hexa-
gon ACNKBH = par. ACNH + par.
HNKB = AH X lN + BH X hL
= sq. HG + sq. HD.

sq. upon AB = sq.
upon BH + sq. upon AH.

a. See Edwards* Geom.,
1895, P. 161, fig. (52); Ver-
sluys, p. 25, fig. 21, cred-
ited to Van Vleth (I805);

also, as an original proof,
by Joseph Zelson a sophomore In West Phlla., Pa.,
High School, 1937.

b. In each of the 39 figures given by Edwards
the author hereof devised the proofs as found herein.

Itlirty-Six

In fig. 136, produce HN to P. Then sq. AK
= (rect. BP = paral. BHNK = sq. HD) + (rect. AP
= paral. HACN = sq. HG).

/. sq. upon AB = sq. upon BH + sq. upon AH.
.*. h^ = a^ + b^.

a. See Math. Mo. (l859), Vol. 2, Dem. 17,

fig. 1.

In fig. 157, the construction Is evident.

Sq. AK = rect. BL + rect. AL = paral. BM + paral. AM
= paral. BN + paral. AO = sq. BE + sq. AP.

sq. upon AB = sq. upon BH + sq. upon AH.
a. See Edwards* Geom., 1895, p. 160, fig.
(28); Ebene Geometrle von G. Mahler, Leipzig, I897,

GEOMETRIC PROOFS

123

p. 80, fig. 60; and Math.

Mo., V. IV, 1897, P. 168,
proof XXXrV; Versluys, p. 37 9
fig. 60, vhere It is credit-
ed to Hauff’s work, I803.

Fig. 137

In fig. 138, the con-
struction is evident, as well
as the parts containing like
numerals.

Sq. AK = trl. BAL
+ trl. CKN + sq. LN + (trl. ACM
+ trl. KBP) + trl. HQA + trl.
OHS + sq. RP + (rect. HL = sq.
HP '+ rect. AP = sq. HD + rect,
GR) = sq. HD + sq. HG.

sq. upon AB = sq.
upon BH + sq. upon AH.

a. See Heath's Math.
Monographs, No. 2, p. 33, proof
XXI.

IlliLil-liiQi

Produce CA to P, draw PHN, join NE, draw HO
perp. to CR, CM par. to AH, join MK and MA and pro-
duce DB to L. Prom this dissection there results:

Sq. AK = rect. AO + rect. BO = (2 trl. MAC = 2 trl.
ACM = 2 trl. HAM = 2 trl. AHP = sq. HG) + (rect. BHMK
= 2 trl. NHL = 2 trl. HLN = 2 trl. NEH = sq. HD).

124

THE PYTHAGOREAN PROPOSITION

Fig. 139

/. sq. -upon AB = sq.

upon HB + sq. upon HA. /.

= + b^. Q.E.D.

a. Devised by the

author Nov. l6, 1933.

Fig. l40 suggests
Its construction, as all
lines drawn are either perp.
or par. to a side of the giv-
en trl. ABH. Then we have
sq. AK = rect. BL + rect. AL
= paral. BHMK + paral. AHMC
= paral. BHNP + paral. AHNO
= sq. HD + sq. HG. /. sq.
upon AB = sq. upon BH + sq.
upon AH.

a. This Is known as
Haynes* proof; see Math. Mag-
azine, Vol. I, 1882, p. 25,
and School Visitor, V. IX,
1888, p. 5, proof IV; also
see Pourrey, p. 72, fig. a.

In Edition arabe des Ele-
ments d’Euclldes.

Farty.-Qne

Draw BQ perp. to AB meeting GP extended, HN
par. to BQ, NP par. to HP, thus forming OARQ; draw
OL par. to AB, CM par. to AH, AS and KT perp. to CM,
and SU par. to AB, thus dissecting sq. AK Into parts
1, 2, 5, 4 and 5.

Sq. AK = paral. AEQO, for sq. AK = [(quad.
ASMB = quad. AHLO ) + (trl. CSA = trl. NPH = trl. OGH)
+ (trl. SUT = trl. OLP) = sq, HG] + [(trap, CKUS

GEOMETRIC PROOFS

125

= trap. NHRP = trl. NVW
+ trap. EWVH, since trl. EPR
= trl. WNV = trap. BDER)

+ (trl. NPQ - trl. HBR) = sq
HD ] = sq. HG + sq. HD.

sq. upon AB = sq.
upon BH + sq. upon HA.

= a + D .

a. This proof and
fig. was formulated by the
author Dec. 12, 1935, to
show that, having given a
paral. and a sq. of equal
areas, and dimensions of
paral. = those of the sq.,
the paral. can be dissected
Into parts, each equivalent
to a like part In the square

For

The construction of
fig. 142 Is easily seen.

Sq. AK = rect. BL -f rect. AL
= paral. HBMN + paral. AHNO
= sq. HD + sq. HG. /. sq.
upon AB = sq. upon BH + sq.
upon AH. h^ = a^ + b^.

a. This Is Lecchlo*s
proof, 1755. Also see Math.
Mag., 1859, Vol. 2, No. 2,
Dem. 3^ and credited to
Charles Young, Hudson, 0.,
(afterwards Prof. Astronomy,
Princeton College, N. J. );
Jury Wlpper, I88O, p. 26,
fig. 22 (Historical Note);
01ney*s Geom. , I872, Part III, p. 251, 5th method;
Jour, of Education, V. XXV, I887, p. 404, fig. Ill;
Hopkins* Plane Geom. , I89I, p. 91> fig. II; Edwards*

126

THE PYTHAGOREAN PROPOSITION

Geom., 1895, P. 159, fig. (25); Am. Math. Mo., V. IV,
1897 f P. 169# XL; Heath’s Math. Monographs, No. 1,
1900, p. 22, proof VI; Versluys, 191^, p. I8, fig.

14.

b. One reference says: "This proof is but a
particular case of Pappus* Theorem."

c. Pappus was a Greek Mathematician of Alex-
andria, Egypt, supposed to have lived between 300 and
400 A.D.

d. Theorem of Pappus: "If upon any two sides
of triangle, parallelograms are constructed, (see
fig. 143), their sum equals the possible resulting
parallelogram determined upon the third side of the
triangle . "

e. See Chauvenet’s Elem’y Geom. (189O), p.
147, Theorem 17. Also see P. C. Boon's proof, 8a,
p. 106.

f. Therefore the so-called Pythagorean Propo-
sition is only a particular case of the theorem of
Pappus; see fig. 144 herein.

Theorem of Pappus

Let ABH be any triangle; upon BH and AH con-
struct any two dissimilar parallelograms BE and HG;

produce GP and DE to C, their
point of intersection; Join C and
H and produce CH to L making KL
= CH; through A and B draw MA to
N making AN = CH, and OB to P mak-
ing BP = CH.

Since trl. GAM = tri. PHC,
being equiangular and side GA
= PH. MA = CH = AN; also BO
= CH = BP = KL. Paral. EHBD
+ paral. HPGA = paral. CHBO
+ paral. HCMA = paral. KLBP
+ paral. ANLK = paral. AP.

Also paral. HD + paral. HG
= paral. MB, as paral. MB = paral.
AP.

GEOMETRIC PROOFS

127

a. As paral. HD and paral. HG are not similar,
it follows that BH® + HA* ^ AB*.

b. See Math. Mo. { 1858 ), Vol. I, p. 558 , Dem.
8, and Vol. II, pp. 45-52, in which this theorem is
given by Prof. Charles A. Yoking, Hudson, 0., now As-
tronomer, Princeton, N.J. Also David E. Smith's
Hist, of Math., Vol. I, pp. 136-7 .

c. Also see Masonic Grand Lodge Bulletin, of
Iowa, Vol. 30 ( 1929 ), No. 2, p. 44, fig.; also Pour-
rey, p. 101, Pappus, Collection, TV, 4th century,

A.D.j also see p. I 05 , proof 8 , in "A Companion to
Elementary School Mathematics," (1924), by P. C . Boon,
A.B.; also Dr. Leltzmann, p. 31 , fig. 32 , 4th Edition;
also Heath, History, II, 355.

d. See '’Companion to Elementary School Mathe-
matlcs," by P. C. Boon, A.B. (1924), p. l4; Pappus
lived at Alexandria about A.D. 300, though date Is un-
certain.

e. This Theorem of Pappus Is a generalization
of the Pythagorean Theorem. Therefore the Pythagorean
Theorem Is only a corollary of the Theorem of Pappus.

EsiLtl-ItlLM

By theorem of Pappus,
MN = LH. Since angle BHA Is
a rt. angle, HD and HG are
rectangular, and assumed
squares (Euclid, Book I, Prop.
47 ). But by Theorem of Pap-
pus, paral. HD + paral. HG
= paral. AK.

sq. upon AB = sq.
upon BH + sq. upon HG. /. h^

= a 4- b .

a. By the author,

Oct. 26 , 1933 .

Pig. 144

128

THE PYTHAGOREAN PROPOSITION

E.£i:i£-F2!ir

Produce DE to L mak-
ing EL = HP, produce KB to 0,
and draw LN perp. to CK. Sq.
AK = rect. MK + rect. MC
= [rect. BL (as LH = MN)

= sq. HD] + (similarly, sq.
HG).

sq. upon AB = sq.
upon HB + sq. upon HG.

= a^ +

a. See Versluys, p.
19, fig. 15> where credited
to Naslr-Ed-Dln (1201-12?^);
also Pourrey, p. 72, fig. 9.

Fgrt^-Fi.V£

In fig. 146 extend
DE and GP to P, CA and KB to
Q and R respectively, draw CN
par. to AH and draw PL and
KM perp. to AB and CN respec-
tively. Take ES = HO and
draw DS.

Sq. AK =

+ hexagon ACKMNB
+ pentagon ACNBH
+ pentagon ftAORP
+ paral. AHPQ +

= sq. HG + trl.

tri. KNM
= trl. BOH
= tri. DSE
:= trl. DES
quad. PHOR
DES + paral.

HG + tri.
sq. HG

/. sq. upon AB =
a. See Am. Math,
proof XLV.

BP - trl. BOH = sq
DES + trap. HBDS =

+ sq. HD.

sq. upon BH + sq. upon AH.
Mo., V. IV, 1897, P. 170,

GEOMETRIC PROOFS

129

Efttii-Six

The construction
needs no explanation; from it
we get sq. AK + 2 tri. ABH
= hexagon ACLKBH = 2 quad.
ACLH = 2 quad, FEDG = hexagon
ABDEPG = sq. HD + sq. HA + 2
tri. ABH.

upon AB = sq.

upon BH +
= a^ +

sq.

sq. upon AH.

h*

a. According to F, C.
Boon, A.B. (1924), p. 107 of
his "Miscellaneous Mathemat-
ics,” this proof is that of
Leonardo da Vinci (1452-1519).

b. See Jury Wipper,
1880, p. 52, fig. 29, as

found in "Aufangsgr unden der Geometrie” von Tempel-
hoff, 1769; Versluys, p. 56, fig, 59, where Tempel-
hoff, 1769, is mentioned; Fourrey, p. 74. Also proof
9, p. 107, in ”A Companion to Elementary School Mathe-
matics,” by F. C, Boon, A.B.; also Dr, Leltzmann,
p. 18, fig. 22, 4th Edition.

In fig. 148 take BO
= AH and AN = BH, and com-
plete the figure; Sq, AK
= rect. BL + rect. AL = paral.
HMKB + paral. ACMH = paral.
PODE + paral. GNEP = sq. DH
+ sq. GH.

/. sq. upon AB = sq.
upon BH + sq. upon AH. /. h^

= a^ + b^.

a. See Edwards* Geom.,
1895, P. 158, fig. (21), and

130

THE PYTHAGOREAN PROPOSITION

Am. Math. Mo., V. IV, I897, p. 169 proof XLI.

I

t

I

Cl

Fig. 149

In fig. 149 extend
CA to Q and complete sq. Q 3 .
Draw GM and DP each par. to
AB, and draw NO perp. to BF.
This construction gives sq.

AB = sq. AN = rect. AL + rect.
PN = paral. BDRA 4 - (rect. AM
= paral. GABO) = sq. HD + sq.
HG.

sq. upon AB = sq.
upon BH + sq. upon AH. h^

= a + D .

a. See Edvards* Geom.,
1895, P. 158, fig. ( 29 ), and
Am. Math. Mo., V. IV, I897,
p. 168, proof XXXV.

EfttilrUine

In fig. 150 extend
KB to meet DE produced at P,
drav QN par. to DE, NO par.
to BP, GR and HT par. to AB,
extend CA to S, draw HL par.
to AC, CV par. to AH, KV and
MU par. to BH, MX par. to AH,
extend GA to W, DB to U, and
drav AR and AV. Then ve will
have sq. AK = trl. ACW + tri.
CVL + quad. AWVY + tri. VKL
+ trl. KMX + trap. UVXM
+ trl. MBU + trl. BUY = (tri.
GRP + trl. AGS + quad. AHRS)

+ (trl. BHT + trl. OND + trap.

HQT) = sq. BE + sq. AP.

GEOMETRIC PROOFS

131

/. sq. upon AB = sq. upon BH + sq. upon AH.

a. This Is E. von Llttrow's proof, 1859; see
also Am. Math. Mo., V. IV, 1897, p. I69, proof XXXVII.

Eifiz

Extend GP and DE to
P, draw PL perp. to CK, ON
par. to AH meeting HB extend-
ed, and KO perp. to AH. Then
there results: sq. AK
[(trap. ACNH - trl, MNH
= paral. ACMH = rect. AL)

= (trap, AHPG - trl. HPP
= sq. AG)] + [ (trap. HOKB

- trl. OMH = paral. HMKB
= rect. BL) = (trap. HBDP

- trl. HEP = sq. HD}].

/. sq. upon AB = sq.
upon BH + sq. upon AH, h^
= a^ + b^.

a. See Am. Math. Mo.,
V. IV, 1897, p. 169, proof
XLII.

F j_f t^-gne

Extend GA to M making
AM = AH, complete sq. HM,
draw HL perp, to CK, draw CM
par. to AH, and KN par. to BH;
this construction gives: sq.
AK = rect. BL + rect. AL
= paral. HK + paral, HACN
= sq. BP + sq. HM = sq. HD
+ sq. HG.

/. sq. upon AB = sq.

upon BH + sq, upon AH. h^

, -^2
= a + b ,

Fig. 152

132

THE PYTHAGOREAN PROPOSITION

a. Vleth*s proof — see Jury Nipper, l880, p,
24, fig. 19 , as given by Vieth, in "Aufangsgrunden
der Mathematik, " 1805; also Am. Math. Mo., V. IV,
1897, P. 169, proof XXXVI.

EitlY-Iwa

FlS. 153

= a^ + b^. Q.E.D.

In fig. 153 construct
the sq. HT, draw GL, HM, and
PN par. to AB; also KG par.
to BH, OS par. to AB, and
join EP. By analysis we
find that sq. AK = (trap.

CTSO + trl. KRU). + [trl. OKU
+ quad. STRQ + (tri. SON
= trl. PRQ) + rect. AQ]

= (trap. EHBV + trl. EVD)

+ [trl. GLP + trl. HMA
+ (paral. SB = paral. ML)]

= sq. HD + sq. AP.

sq, upon AB = sq,
upon BH + sq. upon AH. /. h^

a. After three days of analyzing and classl-
fylng solutions based on the A type of figure, the
above dissection occurred to me, July I6, I89O, from
which I devised above proof.

Fif ee

In fig. 154 through
K draw NL par, to AH, extend
HB to L, GA to 0, DB to M,
draw DL and MN par. to BK,
and ON par. to AO.

Sq. AK = hexagon
ACNKBM = paral. CM + paral.
KM = sq, CO + sq. ML = sq.

HD + sq. HG.

sq. upon AB = sq.
upon BH + sq. upon AH.

(SEOMETRIC PROOFS

133

a. See Edwards' Geom. , I895, p. 157, fig.

(16).

EiftiTiFaiLr

Fig. 155

In fig. 155 extend
HB to M making BM = AH, HA to
P making AP = BH, draw CN
and KM each par. to AH, CP
and KO each perp. to AH, and
draw HL perp. to AB. Sq. AK
= rect. BL + rect. AL = paral.
RKBH + paral. CRHA = sq. RM
+ sq. CO = sq. HD + sq. HG.

sq. upon AB = sq.
upon BH + sq. upon AH. h^

= a^ + h®.

a. See Am. Math. Mo.,
V. IV, 1897, P. 169, proof
XLIII.

Fiftr-Flvs

Fig. 156

Extend HA to N making
AN = HB, DB and GA to M,
draw, through C, NO making
CO = BH, and join MO and KO.

Sq. AK = hexagon
ACOKBM = para. COMA + paral.
OKBM = sq. HD + sq. HG.

/. sq. upon AB = sq.
upon BH + sq. upon AH. /. h^

= a + D .

a. This proof Is
credited to C. French, Win-
chester, N.H. See Journal of
Education, V. XXVIII, l 888 ,
p. 17, 25 d proof; Edwards*
Geom., 1895 , P. 159 , fig. ( 26 );
Heath* 3 Math. Monographs, No.
2 , p. 51, proof XVIII.

134

THE PYTHAGOREAN PROPOSITION

Eiltn-Six

Complete the sq*s OP
and HM, which are equal.

Sq. AK = sq. LN - 4
trl. ABH = sq. OP - 4 trl.
ABH = sq. HD + sq. HG. sq.
upon AB = sq. upon BH + sq.
upon HA. /. h^ = a^ + b^.

Q .E .D .

a. See Versluys, p.
54, fig. 56, taken from Del-
boeuf*s work, i860; Math. Mo.,
1859, Vol. II, No. 2, Dem. 18,
fig. 8; Pourrey, Curios.

Geom. , p. 82, fig. e, I683.

Fig. 157

f t^-Seven

Complete rect. PE and
construct the trl * s ALC and
KMB, each = trl. ABH.

It Is obvious that
sq. AK = pentagon CKMHL - 3
trl. ABH = pentagon ABDNG
- 3 trl. ABH = sq. HD + sq.
HG. sq. upon AB = sq. upon
HD + sq. upon HA. h^ = a^

+ b^.

a. See Versluys, p. 35 $

fig. 57.

Eiftirliaht

In fig. 159 complete the squares AK, HD and
HG, also the paral's KE, GO, AO, PK and BL. Prom

GEOMETRIC PROOFS

135

these we find that sq. AK
= hexagon ACOKBP = paral.

OPGN - paral. CAGN + paral.
POLD - paral. BKLD = paral,
IDMH - (trl. MAE + trl. LDB)
+ paral. GNHM - (trl. GNA
+ trl HMF) = sq. HD + sq. HG.

sq, upon AB = sq. upon BH
+ sq. upon AH. h^ = a^+ b^.

a. See 01ney*s Geom.,
University Edition, 1872, p.
251, 8th method; Edwards’
Geom., 1895, p. 160, fig.
(30); Math. Mo., Vol. II
1859, No. 2, Dem. I6, fig. 8,
and W. Rupert, I9OO.

Fig. 159

Fiftjl-Nine

In fig. 159, omit lines GN, LD, EM, MP and
MH, then the dissection comes to: sq. AK = hexagon
ANULBP - 2 trl. ANO = paral. PC + paral. PK = sq. HD
+ sq. HG. sq. upon AB »= sq. upon HD + sq. upon HA.
h® = + b^. Q.E.D.

a. See Versluys, p. 66, fig. 70.

lUii

In the figure draw
the dlag’s of the sq’s and
draw HL. By the arguments
established by the dissec-
tion, we have quad. ALBH
= quad. ABMN (see proof, fig.
334 ).

Sq. AK * 2 (quad. AKBH

- trl. ABH) * 2 (quad. ABDG

- trl. ABH = J sq. EB + ^ sq.
PA) = sq. HD + sq. HG. sq.
upon AB = sq. upon HD + sq.
upon HA. h^ * a^ + b®.

156

THE PYTHAGOREAN PROPOSITION

a. See E. Pourrey’s Curios. Geom. , p. 96,

fig. a.

SixiY-Qne

GL and DW are each
perp. to AB, LN par. to HB,

QP and VK par. to BD, GR, DS,
MP, NO and K¥ par. to AB and
ST and RU perp. to AB. Tri.
DKV = tri. BPQ. /. AN = MC.

Sq. AK = rect. AP
+ rect. AO = (paral. ABDS
= sq. HD) + (rect. GU = paral.
GABR = sq. GH). sq. upon
AB = sq, upon HB + sq. upon
HA. h® = a® + Q.E.D,

a. See Versluys, p.
28, fig. 24 — one of Werner *s

coll*n, credited to Dohriner,

Sixt)r-Twq

Constructed and num-
bered as here depicted, it
follows that sq. AK = [(trap.
XORB = trap. SBDT) + (tri.

OPQ = tri. TVD) + (quad. PWKQ
= quad. USTE) = sq. HD]

+ [(tri. ACN = tri. PMH)

+' (tri. CWO = tri. GLP)

+ (quad. ANOX = quad. GAML)

= sq. HGj.

sq. upon AB = sq.
upon BH + sq. upon HA. h^

= a® + b®. Q.E.D.

a. See Versluys, p.
55, fig. 32, as given by
Jacob de Gelder, I806.

GEOMETRIC PROOFS

137

Extend GP
and DE to N, com-
plete the square NQ,
and extend HA to P,

GA to R and HB to L,
Prom these
dissected parts of
the sq. NQ ve see
that sq. AK + (4 trl.
ABH + rect. HM
+ rect. GE + rect.

OA) = sq. NQ = (rect.
PR = sq, HD + 2 trl.
ABH) + (rect. AL
= sq. HG + 2 trl.

ABH) + rect. HM
+ rect. GE + rect,

AO = sq. AK + (4 trl.
ABH + rect. HM
+ rect. GE + rect. OA - 2 trl. ABH - 2 trl. ABH -rect.
HM - rect. GE - rect. OA = sq, HD + sq, HG,

/. sq. AK = sq, HD + sq. HG.

sq. upon AB = sq. upon BH + sq. upon AH.

a. Credited by Hoffmann, in "Der Pytha-
goraische Lehrsatz,” 1821, to Henry Boad, of London,
Eng. See Jury Wlpper, l880, p. l8, fig. 12 j Ver-
aluys, p. 53> fig. 55} also see Dr. Leltzmann, p. 20,
fig. 25.

b. Pig. 165 eii5)loys 4 congruent trlangules,

4 congruent rectangles, 2 congruent small squares,

2 congruent HG squares and sq. AK, if the line TB be
Inserted. Several variations of proof Sixty-Three
may be produced from it, if difference is sought, es-
pecially if certain auxiliary lines are drawn.

Fig. 163

158

THE PYTHAGOREAN PROPOSITION

Si£t]^-£2y.j:

Y' '

\

I

77 ii\ — M

/

\

01 l/^ \L

Fig. 164

= sq. HG) + (paral. HBMN =
sq. upon AB = sq.
.. h = a 4- t) .

In fig. 164,
produce HB to L, HA.
to R meeting CK
prolonged, DE and
GP to 0, CA to P,

ED and PG to AB
prolonged. Draw HN
par. to, and OH
perp. to AB, Ob-
viously sq. AK
= trl. RLH - (trl,
RCA + trl • B^j
+ trl, ABH) = trl.
QMO - (trl. QAP
+ tri. OHD + trl.
ABH) = (paral. PANO
sq. HD).

upon HB + sq. upon HA.

\

a. See Jury Wlpper, I 88 O, p. 30, fig. 28a;
Versluys, p. 57# fig. 6 I; Pourrey, p. 82, Pig. c and
d, by H. Bond, in Geometry, Londres, I 683 and 17!53#
also p. 89 .

Sixt^-Fiye

In fig. 165 extend
HB and CK to L, AB and ED
to M, DE and GP to 0, CA
and KB to P and N respec-
tively and draw PN. Now
observe that sq. AK = (trap.
ACLB - trl. BLK) = [quad.
AMNP = hexagon AHBNOP - (trl,
NMB = trl. BLK) = paral BO
= sq. HD) + (paral. AO = sq.
AP)].

sq. upon AB = sq.
upon BH + sq. upon AH.

GEOMETRIC PROOFS

139

a. Devised by the author, July 7, 1901, but
suggested by fig. 28b, In Jury Wlpper, l880, p. 51.

b. By omitting, from the fig., the sq. AK,
and the trl*s BLK and BMD, an algebraic proof through
the mean proportional Is easily obtained.

Sixtu-Six

In the

construction make
CM = HA = PL, LC
= PP, MK - DE = NQ.

OL = LM and MN
= NO. Then sq. AK
= trl. NLM - (trl.
LCA + tri. CMK
+ trl. KNB) = tri.
LNO - (trl. OPH
+ trl. HAB + trl.
QOH) = paral. PLAH
+ paral. HBNQ = sq.
HG + sq. HD. /. sq.
upon AB = sq. upon
BH + sq. upon HA.

/. h^ = a^ + b^.

Fig. 166

Q.E.D.

a. See Versluys, p.
22, fig. 19, by J. D. Krult-
bosch.

Sixty.-Seven

Make AM = AH, BP
= BH, complete paral. MC and
PK. Extend PG and NM to L,
DE and KB to S, CA to T, OP
to R, and draw MP.

Sq. AK = paral. MC
+ paral. PK = paral. LA
+ paral. RB = sq. GH + sq.
HD.

140

THE PYTHAGOREAN PROPOSITION

sq. upon AB = sq. upon HB + sq. upon HA.

/.

a. Math. Mo. (l859), Vol. II, No. 2, Dem. 19,

fig. 9.

sixti-Eiaht

as their altitudes:

: trl. (PHP = sq. HG

Prom P, the middle
point of AB, draw PL, PM and
PN perp. respectively to CK,
DE and PG, dividing the sq*s
AK, DH and PA Into equal
rect * s .

Draw EP, PE, OH to R,
PP and PC.

Since trl's BHA and
EHP are congruent, EP = AB
= AC. Since PH = PA, the
trl»s PAC, HPE and PHP have
equal bases.

Since trl»s having
equal bases are to each other
trl. (HPE = EHP = sq. HD + 4)

+ 4) = EH : PR. trl. HPE

+ trl. PHP : trl. PHP = (ER + PR = AC) :-PR. -J- sq.

HD + i sq. HG ; trl. PHP = AC : PR. But (trl. PAC
= i sq. AK) : trl. PHP = AC : PR. i sq. HD + J sq.

HG : i sq. AK = trl. PHE ; trl. PHP. i sq. HD
+ i sq. HG = ^ sq. AK.

sq. upon AB = sq. upon HB + sq. upon HA.
h® = a® + b®. Q.E.D.

a. Pig. l68 Is tinlque In that It Is the first
ever devised In which all auxiliary lines and all
triangles used originate at the middle point of the
hypotenuse of the given triangle.

b. It was devised and proved by Hiss Ann
Condlt, a girl, aged l6 years, of Central Junior-
Senior High School, South Bend, Ind., Oct. 1938. This
l6-year-old girl has done what no great mathematician,
Indian, Greek, or modern. Is ever reported to have
done. It should be known as the Ann Condlt Proof.

GEOMETRIC PROOFS

141

Hxtrriiins

A~.

- \uy \

Prolong HB to 0 mak-
ing BO = HA.; complete the
rect. OL; on AC const, trl.
ACM = trl. ABH; on CK const,
trl. CKN = tri. ABH. Join
AN, AK, AO, C£, GD, GE and
PE.

It is obvious that
trl. ACN = tri. ABO = trl.

^ I 'y A ABG = trl. EPG; and since

I • ^'v I ^ trl. DEG = [ J(DE) X (AE = AH

+ HE)] = trl. DBG = [i(DB
= DB) X (BP = AE)] = trl. AKO
= [i(K0 = DE) X (HO = AE)]

= trl. AKN = [|(KN = DE)

(an = AE)], then hexagon
ACHKOB - (trl. CNK + trl.

ABO = trl. ABG = trl. EPG)
trl. GBD = trl. GffiD) - (trl.
ACN + 2 trl. ABO - 2 trl.

CNK = 2 trl. GAB + 2 trl. ABD - 2 trl. ABH = sq. AK
= sq. HG + sq. HD.

sq. upon AB = sq. upon HB + sq. upon HA.

Fig. 169

BOK) = (trl. ACN = trl.

+ (tri. AKN = trl. AKO =
CNK + tri. BOK) = 2 trl.

h® =

a^ + b®. Q.E.D.

a. This fig., and proof. Is original; it was
devised by Joseph Zelson, a jimlor In ¥est Phlla.,
Pa., High School, and sent to me by his uncle, Louis
G. Zelson, a teacher In a college near St. Louis, Mo.,

on May 5, 1959.
mentality.

It shows a high intellect and a fine

b. The proof Sixty-Eight , by a girl of I 6 ,
and the proof Slxty-Nlne . by a boy of I 8 , are evi-
dences that deductive reasoning is not beyond our
youth.

142

THE PYTHAGOREAN PROPOSITION

Seventy

Theorem . -
If upon any con~
ventent lenith,
as AB, three tri-
angles are con-
structed, one
having the angle
opposite AB ob-
tuse, the second
having that angle
right, and the
third having that
opposite angle
acute, and upon
the sides includ-
ing the obtuse,
right and acute
angle squares are
constructed, then
the sum of the
three squares are ^
less than, equal
to, or greater
than the square
constructed upon
AB, according as
the angle is ob-
tuse, right or
acute.

In fig. 170, upon AB as diameter describe the
semicircumference BHA. Since all triangles whose ver-
vertex H* lies within the circumference BHA Is ob-
tuse at H*, all triangles whose vertex H lies on that
circumference Is right at H, and all triangles whose
vertex Kz lies without said circumference Is acute at
Hg, let ABH*, ABH and ABHg be such triangles, and on
sides BH* and AH* complete the squares H*D* and H*G*;
on aides BH and AH complete squares HD and HG; on

GEOMETRIC PROOFS

143

sides BHg and AHp complete squares H2D2 and H2G2. De-
termine the points P*, P and P2 and draw P'H* to L*
making N*L* = P*H*, PH to L making NL = PH, and P2H2
to L2 making N2L2 = P2H2.

Through A draw AC’, AC and AC2; similarly
draw BK*, BK and BK2; complete the parallelograms AK* ,
AK and AK2.

Then the paral. AK* = sq. H’D + sq. H’A* .

(See d under proof Forty-Two , and proof imder fig.
143); the paral. (sq. ) AK = sq. HD + sq. HG; and
paral. AK2 = sq. H2D2 + sq. H2G2.

Now the area of AK* Is less than the area of
AK If (N’L* = P*H* ) Is less than (NL = PH) and the
area of AK2 Is greater than the area of AK If (N^L2
= P2H2) Is greater than (NL = PH).

In fig. 171 construct
rect. FHEP = to the rect.

FHEP In fig. 170, take HF*

= H*F In fig. 170, and com-
plete F*H*E*P'; In like man-
ner construct F2H2E2P2 equal
to same In fig. 170. Since
angle AH*B Is always obtuse,
angle E*H*F* Is always acute,
and the more acute E*H*F* be-
comes, the shorter P*H* be-
comes. Likewise, since angle
AH2B Is always acute, angle
E2H2F2 Is obtuse, and the
more obtuse It becomes the

So first: As the variable acute angle F*H*E*
approaches Its superior limit, 90®, the length H*P*
increases and approaches the length HP; as said vari-
able angle approaches. In degrees. Its Inferior limit,
0®, the length of H’P* decreases and approaches, as
Its Inferior limit, the length of the longer of the
two lines H’A or H’B, P* then coinciding with either
E* or F*, and the distance of P* (now E* or F* ) from
a line drawn through H* parallel to AB, will be the
second dimension of the parallelogram AK* on AB; as

longer P2H2 becomes.

144

THE PYTHAGOREAN PROPOSITION

said angle P*H*E* continues to decrease, H*P* passes
through Its Inferior limit and Increases continually
and approaches Its superior limit co , and the distance
of P* from the parallel line through the correspond-
ing point of H* Increases and again approaches the
length HP.

said distance Is always less than HP and
the parallelogram AK* Is always less than the sq. AK.

And secondly: As the obtuse variable angle
E2H2P2 approaches Its Inferior limit, 90°, the length
of H2P2 decreases and approaches the length of HP;
as said variable angle approaches Its superior limit,
180°, the length of H2P2 Increases and approaches 00
In length, and the distance of P2 from a line through
the corresponding H2 parallel to AB Increases from
the length HP to 00, which distance Is the second
dimension of the parallelogram A2K2 on AB.

the said distance Is always greater than
HP and the parallelogram AK2 Is always greater than
the sq. AK,

the sq. upon AB = the Siam of no other two
squares except the two squares upon HB and HA.

/. the sq. upon AB = the sq. upon BH + the sq.

upon AH.

h^ = a^ + b^, and never a*^ + b'^.
a. This proof and figure was formulated by
the author, Dec. I6, 1935*

B

This type Includes all proofs derived from
the figure In which the square constructed upon the
hypotenuse overlaps the given triangle and the squares
constructed upon the legs as In type A, and the
proofs are based on the principle of equivalency.

GEOMETRIC PROOFS

145

Seventh-fine

/ I

\

A

>

l£

\

1-

j y

Fig. 172

trl.

(4 =

+ tri. ( 2=2
' ) = sq. AD ] .

sq. upon AB = sq.

+ b^.

Pig. 172 gives a par-
ticular proof. In rt. trl.
ABH, legs AH and BH are equal.
Complete sq. AC on AB, over-
lapping the tri, ABH, and ex-
tend AH and BH to C and D, and
there results 4 equal equiva-
lent tri*s 1, 2, 5 and 4.

The sq . AC = trl ’ s
[(1 + 2 + 5 + 4), of which
) = sq. BC, and trl. 5 + trl.

upon BH -f- sq. upon AH,

a. See fig. 73b and fig. 91 herein,
b This proof (better. Illustration), by
Richard . Bell, Feb. 22, 1938. He used only ABCD
of fig. 172 ; also credited to Joseph Houston, a high
school boy of South Bend, Ind. , May I 8 , 1939. He
used the full fig.

Seven ih-Iwfi

\ j I
\

>

Fig. 175

and HD) + (quad. PhOK
CMHP common to sq’s AK and HG) =
.*. sq. upon AB = sq. upon BH + sq
+ b^. Q.E.D.

a. This proof, with fig.
author March 26, 1934, 1 p.m.

Take AL = CP and draw
LM and ON perp, to AH.

Since quad. CMNP
= quad. KCOH, and quad. CNHP
is common to both, then quad.
PHOK = trl. CMN, and we have:
sq. AK = (trl. ALM = trl. CPF
of sq. HG) + (quad. LBHM
= quad. OBDE of sq. HD)

+ (trl. OHB common to sq*s AK
trl. CGA of sq. HG) -I- (quad,
sq. HD + sq. HG.

upon HA.

discovered by the

146

THE PYTHAGOREAN PROPOSITION

Ssyent^^-Three

Fig. 174

Assuming the three
squares constructed, as in
fig. 174 , draw GD--it must
pass through H.

Sq. AK = 2 trap. ABML
= 2 tri. AHL + 2 tri. ABH
+ 2 tri. HBM = 2 tri. AHL
-f 2 (tri. ACG = tri. ALG + tri.
GLC) + 2 tri. HBM = (2 tri.
AHL + 2 tri. ALG) + (2 tri. GLC
= 2 tri. DMB) + 2 tri. HBM

= sq. AP + sq. BE,

/. sq. upon AB = sq. upon BH + sq. upon AH.
/. h^ = a^ + h^.

a. See Am. Math. Mo., V. IV, l897> P. 250,

proof XLIX.

SeventxiFour

Take HM = HB, and
draw KL par. to AH and MN par.
to BH.

Sq. AK = tri. ANM
+ trap. MNBH + tri. BKL + tri.
KQL + quad. AHQC = (tri. CQP
+ tri. ACG + quad. AHQC)

+ (trap, RBDE + tri. BRH)

= sq. AP + sq. HD.

sq. upon AB = sq.

upon BH + sq. upon AH. h^ = a^ + h^.

a. See Am. Math. Mo., V, IV, 1897# p. 250,

proof L,

b. If OP is drawn in place of MN, (LO = HB),
the proof is prettier, but same in principle.

c. Also credited to R. A. Bell, Peb. 28,

1958.

GEOMETRIC PROOFS

147

Seven t^-Fj_ve

^ 1 ' _

Fig. 176

In fig. 176, drav GN
and OD par. to AB.

Sq. AK = rect. AQ
+ rect. OK = paral. AD + rect.
AN = sq. BE + paral. AM = sq.
HD + sq. HG.

sq. upon AB = sq.
upon BH + sq. upon AH.

= a® + b®.

a. See Am. Math. Mo.,
V. IV, 1897, P. 250, XLVI.

Seventjt-Six

V

Fig. 177

upon AB

sq.

= a® + b^.

a. See Am. Math. Mo.
proof XLVII; Versluys, 191^#

In fig. 177, draw GN
and DR par. to AB and LM par.
to AH. K is the pt . of Inter-
section of AG and DO.

Sq. AK = rect. AQ
-f rect. ON + rect. LK = (paral.
DA = sq. BE) + (paral. RM
= pentagon RTHMG + trl. CSF)

+ (paral. GMKC = trap. GMSC
+ tr. TRA) = sq. BE + sq. AP.

= sq, upon BH + sq. upon AH.

. V. IV, 1897, P.

p. 12, fig. 7.

250,

Seventy-Seven

In fig. 178, draw LM
through H perp. to AB, and
draw HK and HC.

Sq. AK = rect. LB
+ rect. LA = 2 tri. KHB + 2
trl. CAH = sq. AD + sq. AP.

sq. upon AB = sq.
upon BH + sq, upon AH.

Fig. 178

THE PYTHAGOREAN PROPOSITION

148

/. 4- b^.

a. Versluys, 1914, p. 12, fig. 7; Wlpper,
l880, p, 12, proof V; Edw. Geometry, 1895, P. 159>
fig. 25; Am. Math. Mo., Vol. IV, 1897, P- 250, proof
LXVIII; E. Pourrey, Curiosities of Geometry, 2nd Ed*n,
p. 76, fig. e, credited to Peter Warlns, I762.

+ trl.

=

Draw HL par. to BK, KM
par. to HA, KH and EB.

Sq. AK = (trl. ABH
= trl. ACG) + quad. AHPC com-
mon to sq. AK and sq. -AP
+ (trl. HQM = trl. CPP)+ (trl.
KPM = trl. END) + [paral. QHOK
= 2 (trl. HOK = trl. KHB - trl.
OHB = trl. EHB - trl. OHB
= trl. EOB) = paral. OBNE ]

OHB common to sq. AK and sq. HD.

/. sq. AK = sq. HD + sq. AP.

sq. upon AB = sq. upon BH + sq. upon AH.
a^ + b^.

I 1/1 ^

Fig. 179

a. See Am. Math. Mo., V. IV, l897, p. 250,
proof LI.

b. See Scl. Am. Sup., V. 70, 1910, p. 582,
for a geometric proof, unlike the above proof, but
based upon a similar figure of the B type.

§.6ven t^-N l.ne

Fig. 180

In fig. 180, extend DE
to K, and draw KM perp. to PB.

Sq. AK = (trl. ABH
= trl. ACG) + quad. AHLC com-
mon to sq. AK and sq. AP
+ [(trl. KLM = trl. BNH)

+ trl. BKM = trl. KBD = trap.
BDEN + (trl. KNE = trl. CLP)],
sq. AK = sq. BE + sq. AP.

GEOMETRIC PROOFS

149

/. sq. upon AB = sq. upon BH + sq. upon AH.

+ b^,

a. See Edwards’ Geom. , 1895, p. l6l, fig.
(56); Am. Math. Mo., V. IV, 1897, P. 251, proof LII;
Versluys, 1914, p. 36, fig. 35, credited to Jenny de
Buck.

yiTr

\

1 1.1 I

Fig. 181

upon BH + sq. upon AH.

a. See Am. Math,
proof LVII.

In fig. 181, extend
GP to L making FL = HB and
draw KL and KM respectively
par. to BH and AH.

Sq. AK = (tri. ABH
= tri. CKL " trap. BDEN + tri.
COP) + (tri. BKM = tri. ACG)

+ (tri. KOM = tri. BNH)4-quad.
AHOC common to sq. AK and sq,
HD t sq. HG.

/. sq.

h^ = a^ + b*=

Mo., V. IV, 1897, P. 251,

upon AB = sq,

.2

a. See Edwards’

In fig. 182, extend
DE to L making KL = HN, and
draw ML.

Sq. AK = (tri. ABH
= tri. ACG) + (tri. BMK = i
rect. BL = [trap. BDEN + (tri.
MKL = tri. BNH)] + quad. AHMC
common to sq. AK and sq. AP
= sq. HD + sq, HG.

sq, upon AB = sq.
upon BH + sq. upon AH. h^

= a + b .

Geom,, 1895, P. 158, fig.

(18).

150

THE PYTHAGOREAN PROPOSITION

liaht^-Iwg

\

4 ^

^

In fig. 183, extend GP
and DE to L and draw LH.

Sq. AK = hexagon AHBKLC
paral. HK + paral. HC = sq.

+

HD

+ sq.

upon AB = sq.
upon AH. t

Fig. 183

HG.
sq.

upon BH + sq.

= a® + b®.

a. Original with the
author, July 7> 1901; hut old
for It appears In Olney’ s Geom.,
university edition, I872, p.
250, fig. 37^; Jury Wlpper, I88O, p. 25, fig. 20b, as
given by M. v. Ash, In "Philosophical Transactions,"
1683 j Math. Mo., V. IV, 1897, P. 251, proof LV;
Heath's Math. Monographs, No. 1, 1900, p. 24, proof
IX; Versluys, 1914, p. 55, fig. 58, credited to Henry
Bond. Based on the Theorem of Pappus. Also see Dr.
Leltzraann, p. 21, fig. 25, 4th Edition.

b. By extending LH to AB, an algebraic proof
can be readily devised, thus Increasing the no. of
simple proofs.

In fig. 184, extend GP
and DE to L, and draw LH.

Sq. AK = pentagon ABDLG
- (5 trl. ABH = tri. ABH + reot.
LH) + sq. HD + sq. AF.

sq. upon AB = sq.
upon BH + sq. upon AH. /. h®

= a® + b®.

a. See Journal of Edu-
cation, 1887, V. XXVI, p. 21,
fig. Xj Math. Mo., 1855, Vol.

II, No. 2, Dem. 12, fig. 2.

Fig. 184

GEOMETRIC PROOFS

151

In fig. 185, extend H
drav LM perp. to AB, and draw
HK and HC.

Sq. AK = rect. LB
+ rect. LA = 2 tri. HBK + 2 tri.
AHC = sq. HD + sq. HG.

sq. upon AB = sq.
upon BH + sq. upon AH.

= a^ + b^.

a. See Scl. Am. Sup.,

V. 70, p. 383, Dec. 10, 1910, being No. I6 in A. R.
Colburn* s IO8 proofs; Fourrey, p. Jl, fig. e.

Eiflh

In fig* 186, extend GF
and DE to L, and through H draw
LN, N being the pt. of inter-
section of NH and AB.

Sq. AK = rect. MB
+ rect. MA = paral. HK + paral.
HC =» sq. HD + sq. HG.

/. sq. upon AB = sq.
upon BH + sq. upon AH. /. h^

= a + b ,

a. See Jury Wipper,
1880, p. 13, fig. 5b, andp. 25,
fig. 21, as given by Klagel in "Encyclopaedie, ” I8O8;
Edwards* Geom. , I895, p. 156, fig. (7); Ebene Geome-
trle, von G. Mahler, 1897, P* 87, art. 11; Am. Math.
Mo., V. IV, 1897, P. 251, LIII; Math. Mo., 1859, Vol.
II, No. 2, fig. 2, Dem. 2, pp. 45-52, where credited
to Charles A. Young, Hudson, 0., now Dr. Young, as-
tronomer, Princeton, N.J. This proof is an applica-
tion of Prop. XXXI, Book TV, Davies Legendre; also
Ash, M. V. of Dublin; also Joseph Zelson, Phila., Pa.,
a student in West Chester High School, 1939.

b. This figure will give an algebraic proof.

152

THE PYTHAGOREAN PROPOSITION

Eiflhtz-Six

In fig. 186 It is evident that sq. AK = hex-
agon ABDKCG - 2 trl. BDK = hexagon AHBKLC = (paral.

KH = rect. KN ) + paral. CH = rect. ON) = sq. HD + sq.
HG. sq. upon AB = sq. upon BH + sq. upon AH. h^
= a^ + b^. Q.E.D.

a. See Math. Mo., I 858 , Vol. I, p. 35^, Dem.
8 , where It Is credited to David Trowbridge.

b. This proof Is also based on the Theorem
of Pappus. Also this geometric proof can easily be
converted Into an algebraic proof.

liab-iY-Seven

HEP =
= sq.

(trl. BDK = trap.
HD + sq. HG.

sq. upon AB =

In fig. 187 , extend DE
to K, draw PE, and draw KM par.
to AH.

Sq. AK = (trl. ABH
= trl. ACG) + quad. AHOC com-
mon to sq. AK and sq. AK + trl.
BLH common to sq. AK and sq.

HD + [quad. OHLK = pentagon
OHLPN + (trl. PMK = trl. PLE)

+ (trl. MKN = trl. ONP) = trl.
BDEL + (trl. COP = trl. LEK)]

sq. upon HD + sq. upon HG.
See Am. Math. Mo., V. IV, 1897, P. 251,

/. h^ = a^ + b^. Q.E.D.
a.

proof LVI.

Elfllltl-ELght

In fig, 188 , extend GP and BK to L, and
through H draw MN par. to BK, and draw KM.

Sq. AK = paral. AOLC = paral. HL + paral. HC
= (paral. HK = sq. AD) + sq. HG.

sq. upon AB = sq. upon BH + sq. upon AH.
h^ = a^ + b^.

GEOMETRIC PROOFS

155

Fig. 188

a. See Jury Wipper,
1880, p. 27, fig. 25, vhere
It says that this proof was
given to Joh. Hoffmann, I8OO,
by a friend; also Am. Math.
Mo., 1897, V. IV, p. 251,
proof LIV; Versluys, p. 20,
fig. 16, and p. 21, fig. I8;
Pourrey, p. 73, fig. b.

b. Prom this figure
an algebraic proof is easily
devised.

c. Omit line MN and

we have R. A. Bellas fig. and a proof by congruency
follows. He found it Jan. 3I, 1922.

Eiati

Extend GP to L making
PL = BH, draw KL, and draw CO
par. to PB and KM par. to AH.

Sq. AK = (trl. ABH
= tri. ACG) + trl. CAO common
to .sq*s AK and HG + sq. MH com-
mon to 3q*s AK and HG + [ penta-
gon MNBKC = rect. ML + (sq. NL
= sq. HD)] = sq. HD + sq. HG.

sq. upon AB = sq.

upon BH +
= a 4* b .

sq. upon HA. /.I

Q.E.D.

a. Devised by the author, July 30, 1900,
afterwards found in Pourrey, p. 84, fig. c.

and

Ninety

In fig. 190 produce GP and DE to L, and GA
and DB to M. Sq. AK + 4 trl. ABH = sq. GD = sq. HD
+ sq. HG + (rect. HM = 2 trl. ABH) + (rect. LH = 2
trl. ABH) whence sq. AK = sq. HD -f- sq. HG.

/. sq. upon AB = sq. upon BH + sq. upon AH.
h^ = a^ + b®.

THE PYTHAGOREAN PROPOSITION

154

a. See Jury Wlpper,
1880, p. 17, fig. 10, and is
credited to Henry Boad, as
given by Johann Hoffmann, In
”Der Pythagoralsche Lehrsatz,”
I82I; also see Edwards’ Geom. ,
1895, P. 157, fig. ( 12 ).
Heath’s Math. Monographs, No.
1, 1900, p. 18, fig. 11; also
attributed to Pythagoras, by
W. W. Rouse Ball. Also see
Pythagoras and his Philosophy
In Sect. II, Vol. 10 , p. 259 ,
1904, In proceedings of Royal
Society of Canada, wherein the figure appears as fol-
lows :

MV'

Fig. 190

Look

Fig. 191

Ninetj^-One

Trl’s BAG, MBK, EMC, AEP,
LDH and DLC are each = to trl. ABH.

3q, AM = (sq, KP - 4
tri. ABH) = [(sq. KH + sq. HF
+ 2 rect. GH) - 4 trl. ABH] = sq.
KH + sq. HP.

sq. upon AB = sq. upon
HB + sq. upon HA. h^ = a^ + b®.

a. See P. 0 . Cullen’s
pamphlet, 11 pages, with title.

Fig. 192

GEOMETRIC PROOFS

155

"The Pythagorean Theorem; or a New Method of Demon-
strating it.” Proof as above. Also Fourrey, p. 80,
as the demonstration of Pythagoras according to
Bretschenschneider ; see Simpson, and Elements of Geom-
etry, Paris, 1766.

b. In No. 2, of Vol. I, of Scientia Bac -
calanreus, p. 6I, Dr. Wm. B. Smith, of the Missouri
State University, gave this method of proof as new .
But, see "School Visitor," Vol. II, No. 4, I88I, for
same demonstration by Wm. Hoover, of Athens, 0., as
" adapted from the French of Dalseme .” Also see
"Math. Mo.," 1859, Vol. I, No. 5 , p. 159; also the
same journal, 1859, Vol. II, No. 2, pp. 45-52, where
Prof. John M. Richardson, Collegiate Institute, Bou-
don, Ga., gives a collection of 28 proofs, among
which, p. 47, is the one above, ascribed to Young.

See also Orlando Blanchard* s Arithmetic,

1852, published at Cazenovia, N.Y,, pp. 259-240; also
Thomas Simpson* s "Elements of Geometry," 176O, p. 55,
and p. 51 of his 1821 edition.

Prof. Saradaranjan Ray of India gives it on
pp. 93-94 of Vol. I, of his Geometry, and says it
"is due to the Persian Astronomer Naslr-uddln who
flourished in the 15th century under Jengls Khan,"

Ball, in his "Short History of Mathematics,"
gives same method of proof, p, 24, and thinks it is
probably the one originally offered by Pythagoras.

Also see "Math, Magazine," by Artemas Martin,
LL.D., 1892, Vol. II, No. 6, p. 97. Dr. Martin says:
"Probably no other theorem has received so much at-
tention from Mathematicians or been demonstrated in
so many different ways as this celebrated proposi-
tion, which bears the name of its supposed dlscover-
er .

c. See T. Sundra Row, 1905, p. l4, by paper
folding, "Reader, take two equal squares of paper
and a pair of scissors, and quickly may you know
that AB^ = BH^ + HA^."

Also see Versluys, 1914, his 96 proofs, p. 4l,
fig. 42. The title page of Versluys is:

156

THE PYTHAGOREAN PROPOSITION

ZES EN NEGENTIG BEWIJZEN
Voor Met

THEOREMA VAN PYTHAGORAS
Verzameld en Gerangschikt
Door

J. VERSLUYS
Amsterdam — 1914

Ninet^-Iwg

Fig. 193

In fig. 195, drav KL
par. and equal to BH, through
H draw LM par, to BK, and
draw AD, LB and OH.

Sq. AK = rect. MK
+ rect. MC = (paral. HK = 2
trl. BKL = 2 trl. ABD = sq.
BE) + (2 trl. AHC = sq. AP).

sq, upon AB = sq.
upon BH + sq. upon AH, h^

= a^ + b^.

a. This figure and

proof Is taken from the following work, now In my li-
brary, the title page of which Is shown on the fol-
lowing page.

The figures of this book are all grouped to-
gether at the end of the volume. The above figure
Is numbered 62, and Is constructed for "Proposltlo
XLVII,” In "Llbrum Prlmum, ** which proposition reads,
"In rectangulls trlangulls, quadratum quod a latere
rectum angulum subtedente descrlbltur; aequuale est
els, quae a laterlbus rectimi angulum contlnentlbus
descrlbuntur quadratls . "

GEOMETRIC PROOFS

157

’’Euclides Elementorum Geomet ricorum
Libros Tredecim
Isidorum et Hypsiclem

& Receiitiores de Corporibus Regularibus, &
Procli

Propositiones Geometricas

Claudius Richards

e Societate Jesu Sacerdos, patria Omacensis in libero Comitatu
Burgundae, & Regius Mathematicanim
Professor: dicantique

Philippo mi. Hispaniarum et Indicarum Regi Cathilico.

Antwerpiae,

ex Officina Hiesonymi Verdussii. M.DC.XLV.

Cum Gratia 4c Privilegio"

Then comes the following sentence:

"Proclus in hunc llbrixm, celehrat Pythagoram
Authorem hulus proposltlonis, pro cuius demonstra-
tlone dicltur Dlls Sacrlflcasse hecatomham Taurorum.”
Following this comes the "Supposlto,” then the "Con-
struction” and then the "Demonstration” which con-
densed and translated is: (as per fig. 195) triangle
BKL equals triangle ABD; square BE equals twice tri-
angle ABM and rectangle MK equals twice triangle BKL;
therefore rectangle MK equals square BE. Also square
AG equals twice triangle ARC; rectangle HM equals
twice triangle CAH; therefore square AG equal rectan-
gle HM. But square BK equals rectangle KM plus rec-
tangle CM. Therefore square BK equals square AG plus
square BD.

THE PYTHAGOREAN PROPOSITION

The work from which the above is taken is a
book of 620 pages, 8 Inches by 12 Inches, bound in
vellum, and, though printed in l645 A.D,, is well
preserved. It once had a place in the Sunderland Li-
brary, Blenheim Palace, England, as the book plate
shows — on the book plate is printed — "Prom the Sunder-
land Library, Blenheim Palace, Purchased, April,

1882."

The work has 4o8 diagrams, or geometric fig-
ures, is entirely in Latin, and highly embellished.

I found the book in a second-hand bookstore
in Toronto, Canada, and on July 15, I89I, I purchased
it. (E. S. Loomis. )

C

This type Includes all proofs derived from
the figure in which the square constructed upon the
longer leg overlaps the given triangle and the square
upon the hypotenuse.

Proofs by dissection and superposition are
possible, but none were found.

Fig. 194

In fig. 194, extend KB
to L, take GN = BH and draw MN
par. to AH. Sq. AK = quad. AGOB
common to sq*s AK and AP + (tri.
COK = tri. ABH + tri. BLH)

+ (trap. CGNM = trap. BDEL)

+ (tri. AMN = tri. BOP) = sq. HD
+ sq. HG.

sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^ + b^.

a. See Am. Math. Mo., V.
IV, 1897, P. 268, proof LIX.

b. In fig. 194, omit MN

and draw KR perp. to OC; then take KS = BL and draw

ST perp. to 00. Then the fig. is that of Richard A.

GEOMETRIC PROOFS

159

Bell, of Cleveland, 0., devised July 1, 1918, and
given to me Feb. 28, 1938, along with 40 other proofs
through dissection, and all derivation of proofs by
Mr. Bell (who knows practically nothing as to Eucli-
dian Geometry) are found therein and credited to him,
on March 2, 1938. He made no use of equivalency.

fi.Illitlz.Eour

In fig. 195 f draw DL par.
to AB, through G draw PQ par. to
CK, take GN = BH, draw ON par.

AH and LM perp. to AB.

Sq. AK = quad. AGRB com-
mon to sq*s AK and AF + (trl. ANO
= trl. BRF) + (quad. OPGN = quad.
LMBS) + (rect. PK = paral. ABDL
= sq. BE) + (trl. GRQ = trl. AML)
= sq. BE + sq. AF.

sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^ + b^.

a. Devised by the author,
July 20, 1900.

Ninety-Five

In fig. 196, through G
and D draw MN and DL each par. to
AB, and draw GB.

Sq. AK = rect. MK + rect.
MB = paral. AD + 2 trl. BAG = sq.
BE + sq. AF.

sq, upon AB = sq, upon
BH + sq. upon AH. h^ = a^ + b^.

a. See Am. Math. Mo., V.
IV, 1897, P. 268, proof LXII.

Fig. 196

160

THE PYTHAGOREAN PROPOSITION

HinetY-Six

Fig. 198

\

In fig. 197^ extend PG
to G, drav EB, and through C
draw HN, and draw DL par. to AB.

Sq. AK = 2[quad. ACNM
= (trl. CGN = trl, DBL) + trl.
AGM common to sq. AK and AF
+ (tri. ACG = trl. ABH = trl, AMH
+ tri. ELD)] = 2 trl. AGH + 2
tri. BDE = sq. HD + sq. HG.

sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = b^.

a. See Am. Math. Mo., V.
IV, 1897, P. 268, proof LXIII.

In fig. 198, extend PG
to C, draw HL par. to AC, and
draw AD and HK. Sq. AK = rect.
BL + rect. AL = (2 tri. KBH = 2
tri. ABD + paral. ACMH) = sq. BE
+ sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^+ b^.

a. See Jury Wipper, l880,
p. 11, II; Am. Math. Mo., V. IV,
1897, P. 267, proof LVIII; Pour-
P. 70, fig. b; Dr. Leitz-
mann*s work (1920), p. JO, fig.
51.

In fig. 199, through G draw MN par. to AB,
draw HL perp. to CK, and draw AD, HK and BG.

Sq. AK = rect. MK + rect. AN = (rect. BL = 2
trl. KBH = 2 trl. ABD) + 2 tri. AGB = sq. BE + sq. AP.

GEOMETRIC PROOFS

l6l

A

sq. upon AB = sq. upon
BH + sq. upon AH. = a^+t)^.

a. See Am. Math. Mo., V.
TV, 1897, P. 268, proof LXI.

Ninety-Nine

In fig. 200, extend PG
to C, draw HL par. to BK, and
draw EP and LK. Sq. AK = quad.
AGMB common to sq*s AK and AP
+ (trl, ACG = trl. ABH) + (trl.
CKL = trap. EHBN + tri. BMP)

+ (trl, KML = trl. END) = sq. HD
+ sq. HG.

sq, upon AB = sq. upon
BH + sq. upon AH. = a®+ b®.

a. See Am. Math. Mo., V.
IV, 1897, P. 268, proof LXIV.

One^Hundrgd

In fig. 201, draw PL par.
to AB, extend PG to C, and draw
EB and PK, Sq. AK = (rect. LK
= 2 trl, CKP = 2 trl. ABE = 2 trl.
ABH + trl. HBE = trl. ABH + trl.
PMG + sq. HD) + (rect. AN=paral.
MB).

sq. upon AB = sq. upon
BH + sq. upon AH. h® = a®+ b®.

a. Se§ Am. Math. Mo., V.
IV, 1897, P. 269, proof LXVII.

Fig. 201

162

THE PYTHAGOREAN PROPOSITION

A

0ne_Hy.ndred_0ne

In fig. 202, extend PG
to C, HB to L, draw KL par. to
AH, and take NO = BH and draw OP
and NK par. to BH.

Sq. AK = quad. AGMB com-
mon to 3q*s AK and AP + (tri. ACG
= trl. ABH) + (trl. CPO = trl.BMP)
+ (trap. PKNO + trl. KMN = sq. NL
= sq. HD) = sq. HD + sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. = a^+ b^.

a. See Edwards* Geom. ,
1895, P. 157, fig. (14).

In fig. 203, extend HB to
L making PL = BH, draw HM perp,
to CK and draw HC and HK.

Sq. AK = rect. BM + rect.
AM = 2 trl. KBH + 2 tri. HAG = sq.
HD + sq. HG.

sq. upon AB = sq, upon
BH + sq. upon AH. /. = a^+ b^.

a. See Edwards' Geom,,

1895, p. I6l, fig. (57).

Q.Q.§..tly.!ldred_Th r ee

Draw HM, LB and EP par.
to BK. Join CG, MB and PD.

Sq. AK = paral. ACNL
= paral. HN + paral. HC = (2 trl.
BHM = 2 tri. DEP = sq. HD) + sq.
HG = sq. HD + sq. HG.

.’. sq. upon AB = sq. upon
BH + sq. upon AH. = a^+ b^.

a. See Am. Math. Mo., V.
IV, 1897 , P. 269 , proof IXIX.

GEOMETRIC PROOFS

165

Fig. 205
a. See Math

LXVIII,

In fig. 205, extend PG
to C, draw KN par. to BH, take
NM = BH, draw ML par. to HB, and
draw MK, KP and BE.

Sq. AK = quad. AGOB com-
mon to sq's AK and AP + (tri. ACG
= trl. ABH) + (trl. CLM = trl.
BOP) + [(trl. LKM = trl. OKP)

+ trl. KON = trl. BEH] + (trl.

MKN = trl. EBB) = (trl. BEH + trl.
EBD) + (quad. AGOB + trl. BOP
+ trl. ABC) = 3 q. HD + sq. HG.

.•. sq. upon AB = sq. upon
BH + sq. upon AH. h® = a^+ h®.
Mo., V. IV, 1897, P. 269, proof

In fig. 206, extend PG
to H, draw HL par. to AC, KL par.
to HB, and draw KG, LB, PD and
EP.

Fig. 206

BH + sq.

upon AH.
a. See Am.

mon to
= trl.
EPD = ■
= trl.
HK = J

Sq. AK = quad. AGLB com-
sq's AK and AP + (trl. ACG
ABH) + (trl. CKG = trl.

sq. HD) + (trl. GKL
BLP) + (trl. BLK = ^paral.
sq. HD) = (5 sq. HD + J sq.

HD) + (quad. AGLB + trl. ABH
+ trl. BLP) = sq. HD + sq. AP.

sq. upon AB = sq. upon
h* = a^ + b^.

Math. Mo., V. rv, 1897, P- 268,

proof LXV.

164

THE PYTHAGOREAN PROPOSITION

a. See Am.
proof LXVI,

In fig. 207, extend PG
to C and N, making PN = BD, KB
to 0, (K being the vertex opp. A
In the sq. CB) draw PD, PE and
PB, and draw HL par. to AC.

Sq. AK = paral. ACMO
= paral. HM + paral. HO = [ (paral.
EHLP = rect. EP) - (paral. EOMP
= 2 trl. EBP = 2 trl. DBP = rect.
DP) = sq. HD] = sq. HD + sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^+b^.

Math. Mo., V. IV, 1897, p. 268,

/. sq. upon AB = sq.
. . h = a + b .

In fig. 208,
through C and K draw NP
and PM par. respectively
to BH and AH, and extend
ED to M, HP to L, AG to
Q, HA to N and PG to C.

Sq. AK + rect. HM
+ 4 trl. ABH = rect. NM
= sq. HD + sq. HG 4- (rect.
NQ = rect. HM) + (rect.

GL = 2 trl. ABH)+ (rect.
BM = 2 trl. ABH).

sq. AK = sq. HD
+ sq. HG. = a^+ b®.

Q . E .D •

upon BH + sq. upon AH.

a. Credited by Joh. Hoffmann, In "Der Pytha-
goralsche Lehrsatz," 1821, to Henry Boad of London;
see Jury Wlpper, I88O, p. 19, fig. 15.

GEOMETRIC PROOFS

165

QQ.®-.!lundrei_Eigh t

by dissection, Dec.

By dissection. Draw HL
par. to AB, CP par. to AH and KO
par. to BH. Number parts as In
figure ,

Whence: sq. AK = parts
[(1 + 2) = (1 + 2) In sq. HD)]

+ parts [(5 + 4 + 5) = (5 + 4 + 5
In sq. HG)] = sq. HD + sq. HG.

sq. upon AB = sq. upon
HD + sq. upon HA. = a^+ b^.

Q .E .D .

to show

1955.

a. Devised by the author
a proof of Type-C figure.

D

This type Includes all proofs derived from
the figure In which the square constructed upon the
shorter leg overlaps the given triangle and the
square upon the hypotenuse.

In fig. 210, extend ED
to K, draw HL perp. to CK and
draw HK.

Sq. AK = rect. BL + rect.
AL = (2 trl. BHK = sq. HD)

+ (sq. HE by Euclid’s proof).

/. sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^ + b^
a. See Jury Wlpper, I88O,
p. 11, fig. 5; Versluys, p. 12,

4 , given by Hoffmann.

Fiff. Pin

166

THE PYTHAGOREAN PROPOSITION

In fig. 211, extend ED
to K, draw CL par. to AH, EM par.
to AB and draw PE.

Sq. AK = (quad. ACLN
= quad. EPGM) + (trl. CKL = trl.
ABH = trap. BHEN + trl. EMA)

+ (tri. KBD = trl. PEH) + trl.
BND common to sq*s AK and HD
= sq. HD + sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. = a^+ b^.

a. See Edwards* Geom.,

1895 , p. 155 , fig. ( 2 ).

&!l&-H!i!ldred_Elgven

u

I

I

I

. '

cl

\

\y

T)

I \

I £>
I ✓

I

\Q'

Fig. 212

In fig. 212, ex-
tend PB and PG to L and M
making BL = AH and GM
= BH, complete the rectan-
gle PO and extend HA to N,
and ED to K.

Sq. AK + rect. MH
+ 4 trl. ABH = rect. PO
= sq. HD + sq. HG + (rect.
NK = rect. MH) + (rect. MA
= 2 trl. ABH) + (rect. DL
= 2 trl. ABH); collecting
we have sq. AK = sq. HD
f sq. HG.

sq. upon AB = sq.
upon BH + sq. upon AH.
h*" = + b®.

a. Credited to Henry Boad by Job. Hoffmann,
1821; see Jury Wlpper, I 88 O, p. 20, fig. 14.

CaSOMETRIC PROOFS

167

fing_Hjindr ed_Iweivg

In fig. 213, extend ED
to K, draw HL par. to AC, and
draw CM.

Sq. AK = rect. BL + rect.
AL = paral. HK + paral. HC = sq.
HD + sq. HO.

sq. upon AB = sq. upon
BH + sq. upon AH. h® = a® + b^.

a. Devised by the author,
Aug. 1, 1900.

Fig. 213

fiae_!lundred_Thir tgen

Fig. 214

In fig. 214, extend ED
to K and Q, draw CL perp. to EK,
extend GA to M, take MN = BH,
draw NO par. to AH, and draw EE.

Sq. AK = (trl. CKL = trl.
PEH) + (trl. KBD = trl. EPQ)

+ (trap. AMLP + trl. AON = rect.
GE) + trl. BPD common to sq's AK
and BE + (trap. CMNO = trap.

BHEP) = sq. HD + sq. HG.

sq. upon AB = sq. upon
BH + sq. upon AH. h® = a®+ b®.

a. Original with the
author, Aug. 1, I9OO.

flQ£_!iun(i!ied_FsurtggQ

Employ fig. 214, numbering the parts as there
numbered, then, at once: sq. AK = simi of 6 parts
[(1 + 2 = sq. HD) + (3 + 4 + 5 + 6 = sq. HG) = sq.

HD + sq. HO].

168

THE PYTHAGOREAN PROPOSITION

sq. -upon AB = sq. upon HB + sq. upon HA.
h® = a® + b^. Q.E.D,

a. Formulated by the author, Dec. 19, 1953.

One Hundred^Fif teen

'nK

Fig. ai5

In fig. 215 » extend HA
to 0 making OA = HB, ED to K, and
join OC, extend BD to P and join
EP. Niimber parts 1 to 11 as In
figure. Now: sq. AK = parts 1
+2+3+4+5j trapezoid EPCK
EK + PC

AG =
+ 11

2

sq

xPD = KDxPD = AHx

3 ) + 11 + 4

l +(6 +

2 + 3) = 1 + (6
11 + 4 + (2 = 7
4 + 7 + 10 = (7

HG = parts 7+4+10
1. Sq. HD = parts 3+6.

.*. sq. AK = 1 + 2 + 3 + 4
5 = 1 + (2 = 6 + 7 + 8 ) + 3 + 4
5 = 1 + (6 + 3 ) + 7 + 8 + 4 + 5
It (6 + 3 ) + (7 + 8 = 11 ) + 4
5 = 1 + (6 + 3 ) + 11 + 4 + 5
+ ( 5 = 2 - 4 , since 5 + 4 +

+ 3 ) + 11 + 4 +
+ 4 + 10 ) - 4 =
+ 4 + 10 + 11 +

2-4=1+ (6 +

1 + (6 + 3 ) + 11
1 ) + (6 + 3 ) = sq

5

5)

HG + sq. HD.

sq. upon AB = sq. upon HB + sq. upon HA.
h^ = a^ + b^. Q.E.D.

a. This figure and proof formulated by Joseph
Zelson, see proof Sixty-Nine , a, fig. I 69 . It came
to me on May 5, 1959.

b. In this proof, as In all proofs received
I omitted the column of "reasons’* for steps of the
demonstration, and reduced the argumentation from
many (in Zelson' s proof over thirty) steps to a com-
pact sequence of essentials, thus leaving. In all
oases, the reader to recast the essentials In the
form as given In our accepted modern texts.

By so doing a saving of as much as 60$^ of
page space results--also hours of time for thinker
and printer.

ISAAC NEWTON
1642-1727

GEOMETRIC PROOFS

169

Fig. 216

In fig. 216, through D
draw LN par. to AB, extend ED to
K, and draw HL and CD.

Sq. AH = (rect. AN
= paral. AD = sq. DH) + (rect. MK
= 2 trl. DCK = sq. GH).

sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^+

a. Contrived by the
author, August 1, I9OO.

b. As In types A, B and
C, many other proofs may be de-
rived from the D type of figure.

E

This type Includes all proofs derived from
the figure In which the squares constructed upon the
hypotenuse and the longer leg overlap the given tri-
angle .

In fig. 217, through H
draw LM par. to KB, and draw GB,
HK and HC.

Sq. AK = rect. LB + rect.
LA = (2 trl, HBK = sq. HD) + (2
trl. CAH = 2 trl, BAG = sq. AF),
sq. upon AB = sq. upon
BH + sq. upon AH. h® = a®+b*,
a. See Jury Wipper, 1880,
p, l4, VI; Edwards' Geom., 1895*
p, 162, fig, (58); Am. Math. Mo.,
V, V, 1898, p. 74, proof LXXV;
Versluys, p. l4, fig. 9» one of

170

THE PYTHAGOREAN PROPOSITION

Hoffmann's collection, l 8 l 8 j Pourrey, p. 7I, fig. gj
Math. Mo., 1859, Vol. II, No. 3 , Dem. 13 , fig. 5 .

Fig. 218

In fig. 218, extend DE
to K and draw DL and CM par. re-
spectively to AB and BH.

Sq. AK = (rect. LB
= paral. AD = sq. BE) + (rect.

LK = paral. CD = trap. CMEK
= trap . AOPB ) + (tri . KDN = trl .
CLM) = sq. BE + sq. AF.

sq. upon AB = sq. upon
BH + sq. upon AH. h® = a®

+ h®.

a. See Am. Math. Mo.,

V. V, 1898, p. 74, LXXIX.

fiQ.&-!iyL(ld!ied_Ningteen

In fig. 219, extend KB
to P, draw CN par. to HB, take
NM = HB, and draw ML par. to AH.

Sq. AK = (quad. NOKC
= quad. GPBA) + (trl. CLM = trl.
BPF) + (trap. ANML = trap. BDEO)
+ trl. ABH common to sq's AK and
AP + trl. BOH common to sq's AK
and HD = sq. HD + sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. h® = a^

+ h*^.

Fig. 219 a. Am. Math. Mo., Vol. V,

1898, p. 74, proof LXXVIIj
School Visitor, Vol. Ill, p. 208 , No. 4 l 0 .

GEOMETRIC PROOFS

171

Fig. 220

In fig. 220, extend DE
to K, GA to L, draw CL par. to
AH, and draw LD and HG.

Sq. AK = 2 [trap. ABNM
= tri. AOH common to sq’s AK
and AP + (trl, AHM = trl. AGO)
+ trl. HBN common to sq*s AK
and HD + (tri. BHO = tri. BDN)]
= sq. HD + sq. AP.

sq. upon AB = sq.
upon BH + sq. upon AH.

a. See Am. Math. Mo.,
Vol. V, 1898 , p. 7^, proof
LXXVI.

&!ie_Hy.!ldred_Twg.nty.-ftne

Extend GP and ED to

\ 0, and complete the rect. MO,

and extend DB to N.

Sq. AK = rect. MO

- (4 trl. ABH + rect. NO)

= {(rect. AL + rect. AO)

- (4 trl. AHB + rect. NO)}

= 2 (rect. AO = rect. AD
+ rect. NO) = (2 rect. AD
+ 2 rect. NO - rect. NO

- 4 trl. ABH) - (2 rect. AD
+ rect. NO - 4 trl. ABH)

= (2 rect. AB + 2 rect. HD
Fig. 221 + rect. NP + rect. BO - 4

trl. ABH) = [rect. AB

+ (rect. AB + rect. NP) + rect. HD + (rect. HD + rect,
BO) - 4 trl. ABH] = 2 trl. ABH + sq. HG + sq. HD
+ 2 trl ABH - 4 trl. ABH) = sq. HD + sq. HG.

sq. upon AB = sq. upon BH + sq. upon AH.
h= = a** + b®.

172

THE PYTHAGOREAN PROPOSITION

a. This formula and conversion Is that of the
author, Dec, 22, 1955, "but the figure Is as given In
Am. Math. Mo., Vol. V, I898, p. where see another
somewhat different proof. No. LXXVIII. But same fig-
ure furnishes;

In fig. 221, extend GP and ED to 0 and com-
plete the rect. MO. Extend DB to N.

Sq. AK = rect. NO + 4 tri. ABH = rect. MO
= sq. HD + sq. AP + rect. BO + [rect. AL = (rect. HN
= 2 tri. ABH) + (sq. HG = 2 tri. ABH + rect. NP)],
which coll'd gives sq. AK = sq. HD + sq. HG.

sq. upon AB = sq. upon BH + sq. upon AH.
h^ = a^ + b^.

a. Credited to Henry Boad by Joh. Hoffmann,
in "Der Pythagoralsche Lehr sat z,” 1821; see Jury Nip-
per, 1880, p. 21, fig. 15.

In fig. 222, draw CL and
KL par. respectively to AH and
BH, and draw through H, LP.

Sq. AK = hexagon AHBKLC
= paral. LB + paral. LA = sq. HD
+ sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^ + b^.

a. Devised by the author,
March 12, 1926.

Rect. LM = [sq. AK = (parts 2 common to sq.

AK and sq. HD + 5 + 4 + 5 common to sq. AK and sq. HG)

GEOMETRIC PROOFS

175

+ parts 6 + (7 + 8 = sq. HG)

+ 9 + 1 + 10 + 11 = [sq. AK = sq.

HG + parts |(6 = 2) + 1 = sq.

HD} + parts (9 + 10 + 11 = 2 trl.
ABN + tri. KPE] = [ (sq. AK = sq.
HD + sq. HG) + (2 trl. ABH
+ tri. KPE)], or reot. LM - (2
tri. ABH + trl. KPE) = [ sq. AK
= sq. HD + sq. HA ].

/. 3q. AK = sq. HD + sq.
sq, upon AB = sq. upon
. h = a

HA.
HD +
+ b"

sq. upon HA.

Q.E.D.

a. Original with the author, June 17 > 1939-

b. See Am. Math. Mo., Vol. V, I 898 , p. 74,
proof LXXVIII for another proof, which Is: (as per
essentials ) :

In fig. 225 , extend CA, HB, DE and CK to M,

N, K and L respectively, and draw MN, LN and CO re-
spectively par. to AB, KB and HB.

Sq. AK + 2 trl. AGM + 5 trl. GNP + trap. AGPB
= rect. CN = sq. HD + sq. HG + 2 trl. AGM + 5 trl,

GNP + trap. COEK, which coll*d gives sq. AK = sq. HD
+ sq. HG.

/. sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = a^+ b^.

In fig. 224, extend KB
and CA respectively to 0 and N,
through H draw LM par. to KB, and
draw GN and MO respectively par.
to AH and BH.

Sq. AK = rect. LB + rect.
LA = paral. BHMO + paral. HANM
= sq. HD + sq. AP.

17 ^

THE PYTHAGOREAN PROPOSITION

/. sq. upon AB = sq. upon BH + sq, upon AH,
h® = + b^.

a. Original with the author, August 1, I9OO,

b. Many other proofs are derivable from this
type of figure.

c. An algebraic proof Is easily obtained from
fig. 224.

P

This type Includes all proofs derived from
the figure In which the squares constructed upon the
hypotenuse and the shorter leg overlap the given tri-
angle .

Seven _

In the fig. 225, draw KM
par. to AH.

Sq. AK = (trl. BKM = trl.
ACG) + (trl. KLM = trl. BND)

+ quad. AHLC common to sq*s AK
and AK + (trl. AWE = trl. CLP)

+ trap. NBHE common to sq^s AK
and EB = sq. HD + sq. HG.

sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = a^ + b^.

a. The Journal of Education, V. XXVIII, I888,
p. 17, 24th proof, credits this proof to J. M, Mc-
Cready, of Black Hawk, Wls.; see Edwards* Geom. ,

1895, p. 89, art. 75; Heath* s Math. Monographs, No, 2,
1900, p. 52, proof XIX; Scientific Review, Feb. I6,
1889, p. 31, fig. 30; R. A. Bell, July 1, 1938, one
of his 40 proofs.

b. By numbering the dissected parts, an obvi-
ous proof is seen.

GEOMETRIC PROOFS

175

In fig. 226 , extend AH
to N making HN = HE, through H
draw LiM par. to BK, and draw BN,
HK and HC.

Sq. AK = rect. LB+rect.
LA = (2 tri, HBK = 2 tri. HBN
= 3q. HD) + (2 tri. CAH = 2 tri,
AHC = sq, HG) = sq, HD + sq, HG.

sq. upon AB = sq. upon
BH = sq, upon AH. h^ = a^

+ b^.

a. Original with the author, August 1, 1900.

b. An algebraic proof may be resolved from
this figure.

c . Other geometric proofs are easily derived
from this form of figure.

fi D.i - !i y. 0. d L§. 1 QL t tl L[l§.

In fig. 227, draw LH
perp. to AB and extend it to
meet produced and draw MB, HK
and HC.

Sq. AK = rect. LB+rect.
LA = (paral, HMBK = 2 tri, MBH
= sq, BE) + (2 tri, CAH = 2 tri,
AHC = sq, AP) = sq, BE + sq, AP,
sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = a^

+ b^.

a. See Jury Wlpper,

1880, p. 14, fig. 7; Versluys,
p. 14, fig. 10; Pourrey, p. 71 > f**

176

THE PYTHAGOREAN PROPOSITION

One_Hundred_Thir

Fig. 228

In fig. 228, extend
GA and BD to M, complete the
square ML, and extend AH to

0 .

Sq. AK + 4 trl. ABH
= sq. LM = sq. HD + sq. HG
+ 5 trl. ABH = (trap. BNOP
-H trl. ARE = trl. ABH), which
collected gives sq, AK = sq.
HD + sq, HG.

/. sq. upon AB = sq.
upon BH + sq. upon AH. /. h^

= a + D ,

a. See Jury Wlpper, l880, p. 17, fig. 11,
where It Is credited, by Johann Hoffmann, In "Der

Pythagoralsche Lehrsatz,” 1821, to Henry Boad.

G

This type Includes all proofs derived from
figures In which the squares constructed upon the two
legs overlap the given triangle.

fili-Hjindred^Ihir t^-2ne

In fig. 229, extend PG to C
and draw KL par. to AG.

Sq. AK = quad. AGMB common
to sq*s AK and AP + (trl. ACG = trl.
ABH) + (trl. CKL = trap. NBHE + trl.
BMP) + (trl. KML = trl. BND) = sq.
HD + sq. HG.

sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = a^ + b^.

Q . E .D .

a. See Edwards* Geom. , 1895,
p. l6l, fig. (53); Am. Math. Mo.,

Fig. 229

GEOMETRIC PROOFS

177

V. V, 1898, p. 73, proof LXX; A. R. Bell, Feb. 24,
1938.

b. In Sol. Am. Sup., V. 70, p. 559, Dec. 5,
1910, Is a proof by A. R. Colburn, by use of above
figure, but the argument is not that given above.

I \

I

1 V ' ■

Ckl '^K

In fig. 250, extend FG to C
and ED to K.

Sq. AK = (tri. ACG = trl.
ABH of sq. HG) + (tri. CKL = trap.
NBHE + trl. BMP) + (tri. KBD = tri.
BDN of sq. HD + trap. LMBD common
to sq*s AK and HG) + pentagon AGLDB

common to sq*s AK and HG) = sq. HD
+ sq. HG.

Fig. 250 /. sq. upon AB =

+ sq. upon AH. /. h^
a. See Edwards* Geom. , 1895, P. 159,

(24); Scl. Am. Sup., V. 70, p. 582, Dec. 10,
for a proof by A. R. Colburn on same form of

sq.
= a^ +

upon BH

b^.

fig.

1910,

figure .

fin® -tly.!ldred_IhirtY.- Three

The construction is obvious.
Also that m + n = o + p; also that
trl. ABH and tri. ACG are congruent.
Then sq. AK = 4o + 4p H-q = 2(o + p)
+ 2(0 + p) + q = 2 (m + n) +2(o+ p)
+ q= 2(m+o) + (m+2n + o + 2p
+ q) = sq. HD + sq. HA.

.% sq. upon AB = sq. upon HD
+ sq. upon HA. h^ = a® + b^.
Q.E.D.

Fig. 231 a. See Versluys, p. 48, fig.

49, where credited to R. Joan,
Nepomucen Reichenberger, Phllosophla et Ma the sis
Unlversa, Regensburg, 1774.

b. By using congruent tri*s and trap*s the
algebraic appearance will vanish.

178

THE PYTHAGOREAN PROPOSITION

Fig. 232

Having the construction,
and the parts symbolized, it is evi-
dent that: sq. AK=3o+p+r+s
= (50 + p) + (o + P = s) + r
= 2(o+p) + 2o + r= (in+o) + (m
+ 2n + o + r) = sq,. HD + sq. HG,

sq. upon AB = sq. upon HD
+ sq. upon HA. h^ = a^ 4- b^.

a. See Versluys, p. 48, fig.

50 ; Pourrey, p. 86.

b. By expressing the dimensions of m, n, o,
p, r and s in terras of a, b, and h an algebraic proof
results.

!lQ.§.-tlundr£4.Thirty.-Five

Complete the three sq's AK,
HG and HD, draw CG, KN, and HL
through G. Then

Sq. AK = 2[trap. ACLM = tri.
GMA common to sq’s AK and AP + (tri.
ACG = tri. AMH of sq. AP + tri. HMB
of sq. HD) + (tri. CLG = tri. BMD
of sq. HD)] = sq. HD + sq. HG. h^
= a + b .

Fig. 235 sq. upon AB = sq. upon BH

+ sq. upon AH.

a. See Am. Math. Mo., V. V, I898, p. 73,
proof LXXII.

Draw CL and LK par. respectively to HB and
HA, and draw HL.

Sq. AK = hexagon ACLKBH - 2 tri. ABH = 2 quad.
ACLH - 2 tri. ABH = 2 tri. ACG + (2 tri. CLG = sq. HD)

GEOMETRIC PROOFS

179

+ (2 trl. AGH = aq. HG) - 2 tri. ABH
= aq. HD + aq, HG + (2 tri. ACG = 2
trl. ABH - 2 trl. ABH = aq. HD - aq.
HG.

sq. upon AB = sq. upon HD
+ sq. upon HA. + b^.

Q.E.D.

a. Original by author Oct,
25, 1933.

Fig. 234

ft 0.® - tl iLQ.1 -Itl IL t Ii. S e y e n

1 '' vr '"F

Cki _ _

In fig. 235 , extend FG to C,
ED to K and draw HL par. to BK.

Sq. AK = rect. BL + rect. AL
= (paral. MKBH = sq, HD) + (paral,
CMHA = sq. HG) = sq. HD + sq. HG.

/. sq. upon AB = sq, upon BH
+ sq. upon AH. = a^ + b^.

Q.E.D.

a\ Journal of Education, V.
XXVII, 1888 , p. 327 , fifteenth proof
Fig. 235 by M. Dickinson, Winchester, N.H.;

Edwards* Geom. , 1895, p. 158, fig.
(22); Am. Math. Mo., V. V, I 898 , p. 73, proof LXXI;
Heath* s Math. Monographs, No. 2, p. 28, proof XIV;
Versluys, p. 13, fig. 8 — also p. 20, fig. 17, for
same figure, but a somewhat different proof, a proof
credited to Jacob Gelder, I 8 IO; Math. Mo., 1859, Vol.
II, No. 2, Dem. 11; Pourrey, p. JO, fig. d.

b. An algebraic proof Is easily devised from
this figure.

180

THE PYTHAGOREAN PROPOSITION

Cir _

Fig. 236

Draw HL perp. to CK and ex-
tend ED and PG to K and C resp'ly,

Sq. AK = rect. BL + rect. AL
= (trl. MLK = quad. RDSP + trl. PSB)
+ [trl. BDK - (trl. SDM = trl. ONR)

= (trl. BHA - trl. REA) = quad. RBHE]
+ [(trl. CKM = trl. ABH) + (trl. CGA
= trl. MPA) + quad. GMPA ] = trl. RBD
+ quad. RBHE + trl. APH + trl. MEH
+ quad. GMPA = sq. HD + sq. HG.

.-. sq. upon AB = sq. upon BH

+ sq. upon AH. = a^ + b^. Q.E.D.

fig''

a. See Versluys, p. 46, fig's 47 and 48, as
given by M. Rogot, and made known by E. Pourrey In
his "Curiosities of Geometry," on p. 90.

fiD.e_Hy,Q,l!l§.4_I!lLLiY“NiD.i

In fig. 237> extend AG, ED,

BD and PG to M, K, L and C respective

ly.

Sq. AK = 4 trl. ALP + 4 quad.
LCGP + sq. PQ + trl. AOE - (trl. BNE
'' y’f = trl. AOE) = (2 trl. ALP + 5 quad.

. ' f ' -’ft

CU-"-

Fig. 257

LCGP + sq. PQ + trl. AOE = sq. HG)

+ (2 trl. ALP + quad. LCGP - trl. AOE
= sq. HD) = sq. HD + sq. HG.

/. sq. upon AB = sq. upon BH
f sq. upon AH. + b^.

a. See Jury Wlpper, l880, p.
29, fig. 26, as given by Relchenberger, In Phllosoph-
la et Mathesls Unlversa, etc.,” Ratlsbonae, 177^;
Versluys, p. 48, fig. 49; Pourrey, p. 86.

b. Mr. Richard A. Bell, of Cleveland, 0,,
submitted, Feb. 28, 1958, 6 fig’s and proofs of the
type G, all found between Nov. 1920 and Peb. 28, 1938.
Some of his figures are very simple.

GEOMETRIC PROOFS

I8l

fiJlS-Hundred^For

In fig. 258, extend ED and
PG to K and C respectively, draw HL
perp. to OK, and draw HO and HK.

Sq. AK = rect. BL + rect. AL
= (paral. MKBH = 2 trl. KBH = sq. HD)
+ (paral. CMHA = 2 trl. CHA = sq. HG)
= sq. HD + sq. HG.

sq. upon AB = sq. upon BH
4- sq. upon AH. = a^ +

a. See Jury Wipper, l880,
p. 12, fig. 4.

b. This proof is only a vari-

ation of the one preceding.

c. Prom this figure an algebraic proof is ob-

tainable .

fi Q.&. tl y. Q 2 L 0 0.1^

i

(

I

I

Ct-

\

In fig. 259^ extend PG
to C, HP to L making PL = HB, and
draw KL and KM respectively par.
to AH -and BH.

Sq. AK = {[(trl. CKM

= tri. BKL) - trl. BNP = trap.
OBHE] + (trl. KMN = trl. BOD)

= sq. HD) + [tri. ACG = tri. ABH)
+ (trl . BOD + hexagon AGNBDO )

= sq. HG] = sq. HD + sq. HG.

/. sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^+b^.
a. As taken from ”Phllosophla et Mathesis
Unlversa, etc.," Ratlsbonae, 1774, by Relchenberger;
see Jury Vipper, l880, p. 29, fig. 27.

Fig. 239

L>

182

THE PYTHAGOREAN PROPOSITION

fi!ie_H|indr§.d_E.artjt-Iw2

In fig. 240, extend
HE and HA respectively to N
and L, and complete the sq.

HM, and extend ED to K and BG
to C.

Sq. AK = sq. HM - 4
tri. ABH = (sq. PK = sq. HD)

+ sq. HG + (rect. LG = 2 tri.
ABH) + (rect. OM = 2 tri. ABH)
= sq. HD + sq. HG + 4 tri. ABH
- 4 tri. ABH = sq. HD + sq.

HG.

sq. upon AK = sq. upon BH + sq. upon AH.
h® = a^ + b^.

a. Similar to Henry Bead's proof, London,

1755 j see Jury Wipper, l880, p. l6, fig. 9; Am. Math.
Mo., V. V, 1898, p. 74, proof LXXIV.

Fig. 24o

In fig. 24l, extend PG and
ED to 0 and K respectively, drav
PL par. to AB, and draw HD and PK.

Sq. AK = (rect. AN = paral.
MB) + (rect. LK = 2 tri. CKP = 2
tri. CKO + 2 tri. POK = tri. PMG
+ tri. ABH + 2 tri. DBH) = sq. HD
+ sq. HG.

.*. sq. upon AB = sq. upon BH
+ sq. upon AH. .'. h® = a® + b®.

Q.E.D.

a. See Am. Math. Mo., Vol.
V, 1898, p. 74, proof LXXIII.

Pig. 24 i

GEOMETRIC PROOFS

183

Fig. 242

In fig. 242, produce FG to C,
through D and G draw LM and NO par.
to AB, and draw AD and BG.

Sq. AK = rect. NK + rect. AO
= (rect. AM = 2 trl. ADB = sq. HD)

+ (2 trl. GBA = sq. HG) = sq. HD
+ sq. HG.

/. sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = a^ + b^.

a. This is No. 15 of A. R.
Colburn’s IO8 proofs; see his proof
in Sci. Am. Sup., V. 70, p. 383,

Dec. 10, 1910.

b. An algebraic proof from this figure is

easily obtained.

2 trl. BAD = hx = a®. — (l)

2 trl. BAG = h(h - x) = b®. (2)

(1) + (2) = (5) h® = + b®. (E.S.L.)

Qne_!iiindred_Fartj|f-Fivg

\

In fig. 243, produce
HF and CK to L, ED to K, and
AG to 0, and draw KM and ON
par. to AH.

Sq. AK = paral. AOLB
= [trap. AGFB + (tri. OLM
= tri. ABH) = sq. HG] + (rect.
GN = trl. CLP - (trl. COG
= trl. KLM) - (trl. OLN
= trl. CKP)] = sq. FK = sq.
HDJ = sq. HD + sq. HG.

sq. upon AB = sq.

upon BH + sq. upon AH. .'. h® = a® + b®.

a. This proof Is due to Prln. Geo. M. Phil-
lips, Ph.D., of the West Chester State Normal School,
Pa., 1875; see Heath's Math. Monographs, No. 2, p. 36,
proof XXV.

Pig. 2l)-3

184

THE PYTHAGOREAN PROPOSITION

One.HyLQdlied-EfilllY-Six

Fig. 244

In fig. 244, extend CK
and HF to M, ED to K, and AG to
0 making GO = HB, draw ON par.
to AH, and draw GN.

Sq. AK = paral. ALMB
= paral. GM + paral. AN = [ (tri.
NGO - tri, NPO = trap. RBHE )

+ (tri. KMN = tri. BRD ) ] = sq. HD
+ sq. HG.

sq. upon AB = sq. upon
BH + sq. upon AH. = a^+b^.

a. Devised by the author,
March l4, 1926.

Q.(I§._Hundred_Fgrt}(;-Seven

Fig. 245

Through D draw DR par.
AB meeting HA at M, and through
G draw NO par. to AB meeting HB
at P, and draw HL perp. to AB.

Sq. AK = (rect. NK
= rect. AR = paral. AMDB = sq.
HB) + (rect, AO = paral. AGPB
= sq. HG) = sq. HD + sq. HG.

sq. upon AB = sq.
upon HB + sq. upon HA. h^

= a^ + b^.

a. See Versluys, p. 28, fig. 25. By Werner.

Produce HA and HB to 0 and N resp'ly making
AO = HB and BM = HA, and complete the sq. HL.

Sq. AK = sq. HL - (4 tri. ABH = 2 rect. OG)

= [(sq. GL = sq. HD) + sq. HG + 2 rect. OG] - 2 rect.

OG = sq. HD + sq. HG. sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = + b^.

GEOMETRIC PROOFS

185

a. See Versluys, p.

52 , fig. 5^, as found in Hoff-
mann* s list and in "Des Pytha-
goraische Lehrsatz," l821.

j,ne

Produce CK and HB to
L, AG to M, ED to K, PG to C,
and draw MN and KO par. to AH.

Sq. AK = paral. AMLB
= quad. AGFB -t- rect. GN + (tri,
MLN = tri. ABH) = sq. GH
+ (rect. GN = sq. PO = sq. HD)
= sq. HG + sq. HD. sq. upon
AB = sq. upon HB + sq. upon
HA. h® = a® + b®,

a. By Dr. Geo. M.
Phillips, of West Chester, Pa.,
in 1875 ; Versluys, p. 58 , fig.

62 .

Fig. 24?

H

This type includes all proofs devised from
the figure in which the squares constructed upon the
hypotenuse and the two legs overlap the given tri-
angle .

fina.HuadLei-Eitti

Draw through H, LN perp. to
AB, and draw HK, HC, NB and NA.

Sq. AK = rect. LB + rect.

LA = paral KN + paral. CN = 2 tri.
KHB + 2 tri. NHA = sq. HD + sq. HG.

sq. upon AB = sq. upon HD
+ sq. upon HA. h^ = a^ + , Q.E.D.

a. See Math. Mo., l859> Vol.
II, No. 2, Dem. 15, fig. 7.

186

THE PYTHAGOREAN PROPOSITION

Through H draw LM perp. to
AB. Extend FH to 0 making BO = HP,
draw KO, CH, HN and BG.

Sq. AK = rect. LB + rect.

LA = (2 trl. KHB = 2 trl, BHA = sq.
HD) + (2 trl. CAH = 2 trl. AGB = sq
AP) = sq. HD + sq. AF.

/. sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = a^ + h^.

a. Original with the
Afterwards the first part of
discovered to be the same as
solution in Am. Math. Mo., V
1898, p. 78, proof LXXXI; also see Fourrey, p
fig. h, in his "Curiosities . "

b. This figure gives readily an algebraic

proof.

author .
it was
the
V,

71,

Fig. 250

In fig. 250, extend ED to 0,
draw AO, OB, HK and HC, and draw,
through H, LO perp. to AB, and draw
CM perp. to AH.

Sq. AK = rect. LB + rect. LA
= (paral. HOBK = 2 trl. OBH = sq.

HD) + (paral. CAOH = 2 trl. OHA
= sq. HG) = sq. HD + sq. HG.

sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = a^ + b^.

Q . E . D .

a. See Olney's Geom. , I872,
Part III, p. 251, 6th method; Jour-

nal of Education, V. XXVI, I887,
p. 21, fig. XIII; Hopkins* Geom., I896, p. 9I, fig.

VI; Edw. Geom., 1895, p. I60, fig. (31); Am. Math.

Mo., 1898, Vol. V, p. 74, proof LXXX; Heath* s Math.
Monographs, No. 1, I9OO, p. 26, proof XI.

b. Prom this figure deduce an algebraic proof.

GEOMETRIC PROOFS

187

In fig. 251, draw LM perp.
to AB through H, extend ED to M, and
draw BG, BM, HK and HC.

Sq. AK = rect. LB + rect. LA
= (paral. KHMB = 2 trl. MBH = sq. HD)

+ (2 trl. ARC = 2 tri. AGB = sq. HG)

= sq. HD + sq. HG.

/. sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = a^ + b^.

a. See Jury Wipper, I88O,

p. 15, fig. 8; Versluys, p. 15, fig.

11 .

b. An algebraic proof follows
the "mean prop*l" principle.

One.Hy.CLdred_Fi.fty.-Fgy.r

In fig. 252, extend ED to Q,
BD to R, draw HQ perp. to AB, ON
perp. to AH, KM perp. to CN and ex-
tend BH to L.

S^q. AK = tri. ABH common to
sq*s AK and HG + (trl. BKL = trap.
HEDP of sq. HD + trl. QPD of sq. HG)
+ (tri. KCM = trl. BAR of sq, HG)

+ (trl. CAN = trap. QPBP of sq. HG
+ trl. PBH of sq. HD) + (sq. MN = sq.
RQ) = sq. HD -f sq. HG.

sq. upon AB = sq. upon BH

+ sq. upon AH.

a. See
(13); Am. Math,

h^ =

. -kS

a + b ,

Edwards* Geom, , 1895, P«
Mo., V. V, 1898, p. 7^,

157, fig.

proof LXXXII,

fl[ne_Hyindr§d_Fif tj^-Fiye

In fig. 253, extend ED to P, draw HP, draw
CM perp. to AH, and KL perp. to CM.

188

THE PYTHAGOREAN PROPOSITION

Sq. AK = tri. ANE coimnon to
sq*s AK and NG + trap. ENBH conmion
to sq’3 AK and HD + (trl. BOH = tri.
BND of sq. HD) + (trap. KLMO = trap.
AGPN) + (tri. KCL = tri. PHE of sq.
HG) + (tri. CAM = tri. HPP of sq. HG)
= sq. HD + sq. HG.

sq. upon AB = sq. upon BH
+ sq. upon AH. = a^ +

a. Original with the author,
August 3^ 1890.

h. Many other proofs may he
devised from this type of figure.

In fig. extend

GA to M making AM = AG, GP to
N making PN = BH, complete the
rect. MN, and extend AH and
DB to P and 0 resp*ly and BH
to R.

Sq. AK = rect. MN
(rect. BN + 3 tri. ABH
trap. AGPB) = (sq. HD = sq.
DH) + sq. HG + rect. BN
+ [rect. AL = (rect. HL = 2
tri. ABH) 4 - (sq. AP = tri. ABH
+ trap. AGPB)] = sq. HD + sq.
HG + rect. BN + 2 tri. ABH
+ tri. ABH + trap. AGPB - rect.
tri. ABH - trap. AGPB = sq. HD + sq. HG.

sq. upon AB = sq. upon BH + sq. upon AH.
a^ 4 * h^. Q.E.D.
a. See Jury Nipper, I88O, p. 22, fig. I6,
credited by Joh. Hoffmann in ”Der Pythagoraische
Lehrsatz," l821, to Henry Boad, of London, England.

N

Fig. 25*V

BN

=

GEOMETRIC PROOFS

189

t^-Sev^n

In fig. 255 ve have sq. AK
= parts 1+2+5+4+5+6; sq. HD
= parts ? + 3*; sq. HG = parts 1+4*
+ (7=5)+ (6 = 2); 30 3q. AK(l + 2
+ 3 + 4 + 5 + 6) = sq. HD[2 + 6' = 3)]
+ sq. HG[1 + (4' = 4) + (7 = 5)

+ (2 = 6 )].

sq. upon AB = sq. upon HD
+ sq. upon HA. = a^ + b®. Q.E.D.

a. Richard A. Bell, of Cleve-
land, 0., devised above proof, Nov,
30, 1920 and gave It to me Feb. 28,
1938. He has 2 others, among his

40, like unto It.

I

This type Includes all proofs derived from a
figure In which there has been a translation from Its
normal position of one or more of the constructed
squares .

Symbolizing the hypotenuse- square by h, the
shorter-leg-square by a, and the longer-leg- square
by b, we find, by Inspection, that there are seven
distinct cases possible In this I-type figure, and
that each of the first three cases have four possible
arrangements, each of the second three cases have two
possible arrangements, and the seventh case has but
one arrangement, thus giving I9 sub-types, as follows:

(1) Translation of the h-square, with

(a) The a- and b-squares constructed outwardly.

(b) The a-sq. const *d out*ly and the b-sq. over-
lapping.

(c) The b-sq. const *d out’ly and the a-sq. over-
lapping.

(d) The a- and b-sq*s const *d overlapping.

190

THE PYTHAGOREAN PROPOSITION

(2) Translation of the a-square, vlth

(a) The h- and b-sq*s const *d out’ly.

(h ) The h-sq. const *d out*ly and the b-sq. over-
lapping.

(c ) The b-sq. const *d out*ly and the h-sq. over-
lapping.

(d ) The h- and b-sq*s const 'd overlapping.

(5) Translation of the b-square, with

(a) The h- and a-sq*s const *d out*ly.

(b) The h-sq. const *d out * ly and the a-sq. over-
lapping.

(c ) The a-sq. const*d out*ly and the h-sq. over-
lapping.

(d) The h- and a-sq*s const ’d overlapping.

(4) Translation of the h- and a-sq*s, with

(a) The b-sq. const *d out*ly.

(b) The b-sq. overlapping.

(5) Translation of the h- and b-sq*s with

(a) The a-sq. const *d out’ly.

(b ) The a-sq. const *d overlapping.

(6) Translation of the a- and b-sq's, with

(a) The h-sq. const *d out’ly.

(b) The h-sq. const ’d overlapping.

(7) Translation of all three, h-, a- and b-squares.

Prom the sources of proofs consulted, I dis-
covered that only 8 out of the possible 19 cases had
received consideration. To complete the gap of the
11 missing ones I have devised a proof for each miss-
ing case, as by the Law of Dissection (see fig. Ill,
proof Ten) a proof Is readily produced for any posi-
tion of the squares. Like Agassiz’s student, after
proper observation he found the law, and then the ar-
rangement of parts (scales) produced desired results.

GEOMETRIC PROOFS

191

fine.Hjindrsd.Fiftit-Eiflht

Fig. 256

Case (1), (a).

In fig. 256, the sq.
upon the hypotenuse, hereafter
called the h-sq. has been
translated to the position HK.
Prom P the middle pt. of AB
draw PM making HM = AH; draw
LM, KM, and CM; draw KN = LM,
perp. to LM produced, and CO
= AB, perp, to HM.

Sq. HK = (2 trl. HMC
= HM X CO = sq. AH) + (2 trl.
MLK = ML X KN = sq. BH) = sq.

BH + sq. AH.

sq. upon AB = sq. upon BH + sq. upon AH.
/. h^ = a^ + b^.

a. Original with the author, August 4, I9OO.
Several other proofs from this figure Is possible.

Case (1), (b).

In fig. 257, the
position of the sq's are
evident, as the b-sq.
overlaps and the h-sq. Is
translated to right of
normal position. Draw PM
perp. to AB through B,
take KL = PB, draw LC,
and BN and KO perp. to
LC, and FT perp. to BN.

Sq. BK = (trap.

PCNT = trap. PBDE ) + (trl. CKO = trl. ABH) + (trl.

KLO = trl. BPH) + (quad. BOLQ + trl. BTF = trap. GFBA)
= sq. BH 4- sq. AH.

/. sq. upon AB = sq. upon BH + sq. upon AH.

.*. h^ = a^ + b^.

a. One of my dissection devices.

Fig. 257

192

THE PYTHAGOREAN PROPOSITION

One.Hun^red.SXx t>r

Fig. 258

Case (1), (c).

In fig. 258, draw RA and
produce it to Q, and draw CO, LM
and KN each perp. to RA.

Sq. CK = (trl. COA = trl.
PDB) + (trap. CLMO + trap. PBHE )

+ (trl. NRK = trl. AQG) + (quad.
NKPA + trl. RML = trap. AHPQ)

= sq. HB + sq. CK.

sq. upon AB = sq. upon
BH + sq. upon AH. = a^ + h^.

a. Devised, by author,
to cover Case (l), (c).

Fig. 259

Produce HA to P making
AP = HB, draw PN par. to AB, and
through A draw ON perp. to and
= to AB, complete sq. OL, produce
MO to G and draw HK perp. to AB.

Sq. OL = (rect. AL
= paral. PDBA = sq. HD) + (rect.
AM = paral. ABCG = sq. HG = sq.

HB + sq. HG.

sq. upon AB = sq. upon
HD + sq. upon HA. h^ = a^

+ b®. Q.E.D.

a. See Versluys, p. 27, fig. 23, as found In
"Friend of Wisdom," I887, as given by J. de Gelder,
1810, In Geom. of Van Kunze, l842.

GEOMETRIC PROOFS

193

ft[l«-.!ly.Q.4Le4-SLit^-Iwo

Case (l), (d).

Draw HO perp. to AB and
equal to HA, and KP par. to AB and
equal to HB; draw CN par. to AB,
PL, EP, and extend ED to R and BD

to Q-

= trl. PER of
= sq. HD + sq
sq

Sq. CK = (trl, LKP = trap,
sq. HD + trl. ASE of sq, HG)
HOB = trl, SDB of sq. HD
AQDS of sq. HG) + (trl. CNH
PHE of sq, HG) + (trl, CLT
sq. TO = sq. DG of sq. HG

=

(d).

a + D

a.

ESBH of
+ (trl.

+ trap.

= trl.
sq. HG) +

HG.

upon AB = sq

Q.E.D.

Conceived, by author, to cover case (l).

upon BH + sq. upon AH.

2Q.®-tlSi[ldLSi-Si2Ltii-I!lLce

\

\

\

■ ^ jN

cUl-k-I'iK

Fig. 261

Case (2), (a).

In fig. 261, with
sq*s placed as In the figure,
draw HL perp. to CK, CO and
BN par. to AH, making BN
= BH, and draw KN.

Sq. AK = rect. BL
+ rect. AL = (paral. OKBH
= sq. BD) + (paral. COHA
= sq. AP) = sq. BD + sq. HG.

/. sq. upon AB = sq.
upon BH + sq. upon AH. h^

= a2 + b2.

a. Devised, by author,
to cover Case (2), (a).

19k

THE PYTHAGOREAN PROPOSITION

In fig. 262, the sq. AK
= parts 1+2+5+4+5+6
+ l6. Sq. HD = parts (12 = 5)

+ (15 = -^) of sq. AK. Sq. HG
= parts (9 = 1) + (lO = 2) + (ll
= 6) + (14 = 16) + (15 = 5) of
sq. AK.

/. sq. upon AB = sq. upon
HD + sq. upon HA. = a^ + b^.

Q.E.D.

a. This dissection and
proof is that of Richard A. Bell,
devised by him July 15, 191^^ and
given to me Peb. 28. 1938.

One _Hiindred_$lxtY» Five

Case (2), (b).--Por which are

more proofs extant than for
any other of these 19 cases-
Why? Because of the obvious
dissection of the resulting
figures .

In fig. 265, extend PG to
C. Sq. AK = (pentagon AGMKB
= quad. AGNB common to sq*s AK
and AP + trl. KNM common to sq* s
AK and PK) + (trl. ACG = trl. BNP
+ trap. NKDP) + (trl. CKM = trl. ABH) = sq. PK + sq.
AP.

/. sq. upon AB = sq. upon BH + sq. upon AH.
h^ = a^ + b^.

a. See Hill*s Geom. for Beginners, I886, p.
154, proof I; Beman and Smith* s New Plane and Solid
Geom., 1899, p. 104, fig. 4; Versluys, p. 22, fig. 20
as given by Schlbmllch, 1849; also F. C. Boon, proof
7, p. 105; also Dr. Leltzmann, p. I8, fig. 20; also

Fig. 265

GEOMETRIC PROOFS

195

Joseph Zelson, a 17 year-old boy In West Phlla., Pa.,
High School, 1937.

b. This figure is of special interest as the
sq. MD may occupy 15 other positions having common
vertex with sq. AK and its sides coincident with side
or sides produced of sq. HG. One such solution is
that of fig. 256.

In fig. 264, extend PG to C.
Sq. AK = quad. AGPB common to 3q*3
AK and AP + (tri. ACG = trl. ABH)

+ (trl. CME = tri. BPP) + (trap.

EMKD common to sq's AK and EK)

+ (trl. KPD = tri. MLK) = sq. DL
+ sq. AF.

sq. upon AB = sq. upon BH
+ sq. upon AH. A h^ = a^ + b^,

a. See Edwards' Geom. , 1895#
p. 161, fig. (35); Dr. Leitzmann,
p. 18, fig. 21. 4th Edition.

2Q.£_HiLndrfed_S j,xt ji- Jevgn

In fig. 265, extend PG to Or
and const, sq, HM = sq. LD, the sq.
translated.

Sq. AK = (trl. ACG = trl.

ABH) + (trl. COE = trl. BPP) + (trap.
EOKL common to both sq's AK and LD,
or = trap. NQBH) + (trl. KPL = tri.
KOD = trl. BQM) + [(trl. BQM 4- poly-
gon AGPBMQ) = quad. AGPB common to
sq's AK and AP] = sq. LD + sq. AP.

/. sq. upon AB = sq. upon BH
+ sq. upon AH, h^ = a^ + b^,

a. See Scl. Am. Sup., V. 70,
p. 359# Dec. 3, I9IO, by A, R. Colburn.

b. I think it better to omit Colburn's sq. HM
(not necessary), and thus reduce It to proof above.

Fig. 265

Fig. 264

196

THE PYTHAGOREAN PROPOSITION

GEOMETRIC PROOFS

197

198

THE PYTHAGOREAN PROPOSITION

several proofs for each is possible. Again, by In-
spection, we observe that the given triangle may have
any one of seven other positions within the square
AGFH, right angles coinciding. Furthermore the
square upon the hypotenuse may be constructed over-
lapping, and for each different supposition as to the
figure there will result several proofs unlike any,
as to dissection, given heretofore.

c. The simplicity and applicability of fig-
ures under Case (2), (b ) makes it worthy of note.

In fig. 271, sq. AK = sec-
tions [5 + (6 = 3)+ (7 = ^)]

+ [(8=1)+ (9 = 2 )]= sq. HG
+ sq. AE.

sq. upon AB = sq. upon
BH + sq, upon HA. h^ = a^ + b^.

Q.E.D.

a. Devised by Richard Bell,
Cleveland, 0., on July 4, 1914,
one of his 40 proofs.

Fig. 272

Case (2), (c).

In fig. 272, ED being the
sq. translated, the construction
Is evident.

Sq. AK = quad. AHLC common
to 3q*3 AK and AP + (trl. ABC
= trl, ACG) + (trl. BKD = trap.
LKEP + trl. CLP) + trl. KLD common
to sq’s AK and ED = sq, ED + sq.
AP.

sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = a^ 4- b^.

GEOMETRIC PROOFS

199

a. See Jury Wlpper, l880, p. 22, fig. 17, as
given by von Hauff, In "Lehrbegrlff der relnen Mathe-
matlk," 1805; Heath * s Math. Monographs, 1900, No. 2,
proof XX; Versluys, p. 29, fig* 27; Pourrey, p. 85 —
A. Marre, from Sanscrit, ”Yonctl Bacha" ; Dr. Leltz-
mann, p. 17, fig. 19, ^th edition.

fta§L.tliiaii:ai.sgvfentit-Five

Fig. 273
a. See Math.

Having completed the
three squares AK, HE and HG, draw,
through H, LM perp. to AB and
join HC, AN and AE.

Sq. AK = [rect. LB
= 2(trl. KHP = trl. AEM) = sq. HD]
+ [rect. LA = 2 (trl. HCA = trl.
ACH) = sq. HG] = sq. HD + sq. HG.

sq. upon AB = sq. upon
HB + sq. upon HA. /. h^ = a^ + b^.
Mo. (1859), Vol. II, No. 2, Dem.

14, fig. 6.

Fig. 274

In fig. 274, since parts
2 + 3 = sq. on BH = sq. DE, It
Is readily seen that the sq. upon
AB = sq. upon BH + sq. upon AH.

/. h^ = a^ + b^.

a. Devised by Richard A.
Bell, July 17, 1918, being one of
his 40 proofs. He submitted a
second dissection proof of same
figure, also his 3 proofs of Dec.
1 and 2, 1920 are similar to the
above, as to figure.

yfri

200

THE PYTHAGOREAN PROPOSITION

fin&-Hl^!ldLed.Sev£rit>r.Sevin

Case (2), (d).

Fig. 275

In fig. 275 ^ extend KB
to P, CA to R, BH to L, draw KM
perp. to BL, take MN = HB, and
drav NO par. to AH.

Sq. AK = trl. ABH common
to sq’s AK and AP + (trl. BON
= trl. BPP) + (trap. NOKM = trap.
DRAE) + (trl. KLM = trl. ARQ)

4 - (quad. AHLC = quad. AGPB ) = sq.
AD + sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. /. = a^ + b^.

a. See Am. Math. Mo., V.
VI, 1899, P. 34, proof XC.

In fig. 276, upon CK
const, trl. CKP = trl. ABH, draw
ON par. to BH, KM par. to AH,
draw ML and through H draw PO.

Sq. AK = rect. KO + rect.
CO = (paral. PB = paral. CL = sq.
AD) + (paral. PA = sq. AP) = sq.
AD + sq. AP.

sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^ + b^.

a. Original with the
author, Jul^r 28, I9OO.

b. An algebraic proof
comes readily from this figure.

Fig. 276

GEOMETRIC PROOFS

201

Case (5), (a).

' \^\ ' '' i

iH r

Fig. 277
= a® +1)^,

In :

^ ' \

1

DB to N, HB

draw CN, AO

iy \

to NB.

Sq.

5| \ 1

+ trl. BRQ =

+ (trl, KBP = tri. TBG)

+ (trap. OQRA = trap. MBDE)

+ (trl. ASO = tri. BMH) = sq.
HD + sq, GL.

sq. upon AB = sq.
upon BH + sq. upon AH. /.

17 , 1926 .

a. Devised for missing Case ( 5 ), (a), March

Case ( 3 ), (b).

In fig. 278, extend ED to
K and through D drav CM par. to
AB,

Sq. AK = rect. AM + rect.
CM = (paral. GB = sq. HD) + (paral.
CD = sq. GP) = sq. HD + sq. GP.

sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = a^ + b^.

a. See Am. Math. Mo.,

Vol. VI, 1899, P. 55 , proof LXXXV.

b. This figure furnishes an algebraic proof,

c. If any of the triangles congruent to trl,
ABH Is taken as the given triangle, a figure express-
ing a different relation of the squares Is obtained,
hence covering some other case of the 19 possible
cases.

Fig. 278

202

THE PYTHAGOREAN PROPOSITION

S''

\

I ' \

' H sF£> '
! \1 .

- ->:k

Extend HA to G making
AG = HB, HB to M making BM
= HA, complete the sqii^re’s
HD, EC, AK and HL. Number the
dissected parts, omitting the
trl*s CLK and KMB.

Sq. (AK = 1 + 4 + 5
+ 6 ) = parts (l common to sq*s
HD and AK) + (4 common to sq*s
EC and AK) + (5 = 2 of sq. HD
^ + 3 of sq. EC ) + (6 = 7 of sq.

EC) = parts (l + 2 ) + parts
/ Fig. 279 (3 + 4 + 7 ) = sq. HD + sq. EC.

sq. upon AB = sq.

upon BH + sq. upon AH. .% h^ = a^ + b^. Q.E.D.

a. See "Geometric Exercises In Paper Folding"
by T. Sundra Row, edited by Beman and Smith (1905),
p. 14 .

\

ani.tiy.n4r£4^Eiaht3t-Iwq

Fig. 280

In fig. 280, extend EP
to K, and HL perp. to CK.

Sq. AK = rect. BL + rect.
AL = paral. BP + paral. AF = sq.
HD + sq. GP.

/. sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^ + b^.

a. See Am. Math. Mo., V.
VI, 1899, P. 33 , proof LXXXIV.

ttD.SL-tly.a4r£4.Eiflht)^-Thrfe§

In fig. 281, extend EP to K.

Sq. AK = quad. ACPL common to sq*s AK and GP

GEOMETRIC PROOFS

203

+ (trl. CKP = trap. LBHE + trl.
ALE) + (trl. KBD = trl. GAG)

+ tri. BDL common to sq*3 AK and
HD = sq. HD + sq. AK.

/. sq. upon AB = sq. upon
BH + sq. upon AH. = a^ + b^.

a. See Olney's Geom. ,

Part III, 1872, p. 250, 2nd meth-
od; Jury Wlpper, I88O, p. 23, fig.
18; proof by E. Forbes, Winches-
ter, N.H., as given In Jour, of
Ed*n, V. XXVIII, 1888, p. 17, 25th proof; Jour, of
Ed»n, V. XXV, 1887, p. 404, fig. II; Hopkins* Plane
Geom., 1891, p. 91 , fig. Ill; Edwards* Geom., 1895#
p. 155, fig. (5); Math. Mo., V. VI, 1899, p. 53,
proof LXXXIII; Heath’s Math. Monographs, No. 1, I9OO,
p. 21, proof V; Geometric Exercises In Paper Folding,
by T. Sundra Row, fig. 13, p. 14 of 2nd Edition of
The Open Court Pub. Co., 1905. Every teacher of
geometry should use this paper folding proof.

Also see Versluys, p. 29, fig. 26, 3rd para-
graph, Clalraut, 1741, and found In "Yonctl Bacha”;
also Math. Mo., I858, Vol. I, p. I60, Dem. 10, and
p. 46, Vol. II, where credited to Rev. A. D. Wheeler.

b. By dissection an easy proof results. Also
by algebra, as (in fig. 281 ) CKBHG = a^ + b^ + ab;
whence readily h^ = a® + b^.

c. Fig. 280 Is fig. 281 with the extra line
HL; fig. 281 gives a proof by congruency, while fig.
280 gives a proof by equivalency, and It also gives
a proof, by algebra, by the use of the mean propor-
tional.

d. Versluys, p. 20, connects this proof with
Macay; Van Schooter, 1657; J. C. Sturm, 1689; Dobrin-
er; and Clalraut.

Fig. 281

In fig. 282, from the dissection it Is obvi-
ous that the sq. upon AB = sq. upon BH + sq. upon AH.

GEOMETRIC PROOFS

205

/. sq. upon AB = sq. upon BH + sq. upon AH.

/. + b^.

a. See Halsted’s Elements of Geom. , 1895#
p. 78 , theorem XXXVII; Edwards* Geom. , 1895# P. 158#
fig. ( 6 ); Heath* s Math. Monographs, No. 1, 1900, p.
27 , proof XIII.

a!l£«HyLndrid.Eiaht3t-Se^^[l

In fig. 285 It Is ob-
vious that the parts In the
sq. HD and HP are the same
In niimber and congruent to
the parts In the square AK.

the sq. upon AB
= sq. upon BH + sq. upon AH,
or h^ = a^ + b^.

a. One of R. A. Bell's
proofs, of Dec. 5, 1920 and
received Feb. 28, 1958.

Safe-. tit

Case (3), (d).

Fig. 286
LXXXIX.

In fig. 286 , produce AH
to 0, draw CN par. to HB, and ex-
tend CA to G.

Sq. AK = trap. EMBH com-
mon to sq*s AK and HD + (trl. BOH
= trl. BMD) -f (quad. NOKC = quad.
FMAG) + (trl. CAN = trl. GAL)

+ trl. AME common to sq*s AK and
EG = sq. HD + sq. LF.

/. sq. upon AB = sq. upon
BH 4- sq. upon AH. /. h^ = a^ + b^.

a. See Am. Math. Mo.,

Vol. VI, 1899 # P. 5^# proof

b. As the relative position of the given tri-
angle and the translated square may be Indefinitely

206

THE PYTHAGOREAN PROPOSITION

varied, so the niomber of proofs must be indefinitely
great, of which the following two are examples.

Fig, 287

In fig. 287, produce BH
to Q, HA to L and ED to P, and
draw KN perp. to QB and connect A
and G.

Sq, AK = tri. APE common
to sq*s AK and EG + trap. PBHE
common to sq*s HD and AK + (tri.
BKN = tri. GAL) + (tri. NKQ
= tri. DBP) + (quad. AHQC=quad.
GPPA) = sq, HD + sq. HA,

/. sq. upon AB = sq, upon
HD + sq. upon HA. /. h^ = a^ + b^,
a. This fig. and proof
due to R. A, Bell of Cleveland, 0.
He gave it to the author Feb. 27,

1938.

In fig. 288, draw LM
through H.

Sq. AK = rect, KM + rect*
CM = paral. KH + paral. CH = sq.

HD + (sq. on AH = sq. NF),

/. sq. upon AB = sq. upon
BH + sq. upon AH. h^ = a^ + b^.

a. Original with the
author, July 28, I9OO.

b. An algebraic solution
may be devised from this figure.

Fig. 288

GEOMETRIC PROOFS

207

Case ( 4 ), (a).

In fig. 289, extend
KH to T making NT = AH, draw
TC, draw PR, MN and PO perp.
to KH, and draw H 3 par. to
AB.

' Sq. CK = (quad. CMNH
+ trl. KPO = quad. 5 HFG)

+ tri. MKN = trl. HSA)

+ (trap. PROP = trap. EDLB)

+ (trl. PHR = trl. ECB) = sq.
CD + sq. GH.

sq. upon AB = sq.

upon BH + sq. upon AH. h® = a® + b®.

a. Devised by author for case ( 4 ), (a) March

18, 1926.

Fig. 290

Case ( 4 ), (b).

In fig. 290, draw GP
par. to AB, take LS = AH,
draw KS, draw LO, CN and QM
perp. to KS, and draw BR.

Sq. AK = (trl. CNK
= trl. ABH) + (tri. KQM
= tri. PBR) + (trap. QLOM
= trap. PGED) + (trl. SOL
= trl. OPR) + (quad. CNSA
= quad. AGRB) = sq. GD + sq.
AP.

sq. upon AB = sq.

upon BH + sq. upon AH. /. h® = a® + b®.

a. Devised by author for Case ( 4 ), (b).

208

THE PYTHAGOREAN PROPOSITION

ia£-!i!^niii:e.i-!iLae.iYrI!iL&&

\

/. sq. RW = sq

upon AH.

a. Devised March
by author.

Case (5), (a).

In fig. 291, CE and AP
are the translated sq^s; pro-
duce GP to 0 and complete the
sq. MO; produce HE to S and
complete the sq. US; produce
OB to Q, draw MP, draw WH,
draw ST and UV perp, to WH,
and take TX = HB and draw XY
perp. to WH. Since sq. MO
= sq. AP, and sq. US = sq. CE,
and since sq. RW = (quad. URHV
+ trl. WYX = trap. MPOB + (trl.
HST = trl. BQH) + (trap. TSYX
= trap. BDEQ) + trl. UVW
= trl. MPN) = sq. HD + (sq. NB
= sq. AP).

upon AB = sq. upon BH + sq.

18, 1926, for Case (5), (a).

Fig. 292

Extend HA to G mak-
ing AG = HB; extend HB to D
making BD = HA. Complete
sq’s PD and PG, Draw HQ
perp. to CK and through P
draw LM and TU par. to AB.

PR = CO = BW.

The translated sq*s
are PD = BE* and PG = HG* .

Sq. AK = parts (l
+ 2 + 5 + 4 + 5 + 6+ 7 + 8)
= parts (3 + 4 + 5 + 6= sq.
PD) + parts (1 + 2 + 7 + 8)

= sq. PG.

C3E0METRIC PROOFS

209

/. sq. upon AB = sq. upon HB + sq. upon HA.
a® + b®, Q.E.D.

a. See Versluys, p. 55, fig. 54.

Pig. 293

Case (5), (b).

In fig. 293, draw GL
through B, and draw PQ, CO and
MN perp. to BL.

Sq, BK = (trl. CBO = trl.
BGD) + (quad. OCKL + trl. BPO
= trap. GPRB) + (trl. MLN = trl.
BSD) + (trap. PQNM = trap. SEHB)

= sq. HD + sq. DP.

sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = a^ + b^.

a. Devised for Case (5),
(b), by the author, March 28,

1926.

Case ( 6 ), (a).

Fig. 29^1-

In fig. 294 , extend
LE and PG to M thus completing
the sq. HM, and draw DM.

Sq. AK + 4 trl. ABC
= sq. HM = sq. LD + sq. DP
+ (2 rect. HD = 4 trl. ABC),
from which sq. AK = sq. LD
+ sq. DP.

.% sq. upon AB = sq.
upon BH + sq. upon AH. h^

= a^ + b^.

a. This proof Is cred-
ited to M. McIntosh of Whltwater, Wls. See Jour, of
Ed*n, 1888 , Vol. XXYII, p. 327, seventeenth proof.

210

THE PYTHAGOREAN PROPOSITION

rf

Sq. AK = sq. HM - (4

trl. ABH = 2 rect. HL = sq. EL
+ sq. LP + 2 rect. HL - 2 rect.
HL = sq. EL + sq. LP.

sq. upon AB = sq.
upon HB + sq. upon HA.

= + b®.

a. See Journal of Edu-
cation, 1887, Vol. XXVI, p. 21,
fig. XII; Iowa Grand Lodge Bul-
letin, P and A.M., Vol. 50, No.
2, p. 44, fig. 2, of Peb. 1929.
Also Dr. Leitzmann, p. 20, fig.
24, 4th Ed'n.
b. An algebraic proof is h® = (a + b)® - 2ab

X / I '

I

V ’ ^ ' S

V \ ’/'f

Pig. 295

= a® + b‘

In fig. 296, the
translation is evident.
Take CM = KD. Draw AM;
then draw GR, ON and BO
par. to AH and DU par. to
BH. Take HP = BH and
draw PQ par. to AH.

Sq. AK = (tri.

CMN = trl. DEU) + (trap.
CNPQ = trap. TKDU)

+ (quad. OMRB + trl. AQP)
= trap. PGRQ) + trl. AOB
= trl. GCR) = sq. EK + sq. PC.

sq. upon AB = sq. upon HB + sq. upon HA.
h® = a* + b^. Q.E.D.

a. Devised by the author, March 28, 1926-.

7T

r Th4:

> V

\

!

Pig. 296

GEOMETRIC PROOFS

211

!ine_Hu.ndr§.i_NJ[nftt)^-Ni,n£

Fig. 297

In fig. 297, the trans-
lation and construction is ev-
ident .

Sq. AK = (trl. CRP
= trl. BVE) + (trap. ANST
= trap. BMDV ) + (quad. NRKB
+ tri. TSB = trap. APGC ) + trl.
AGP common to sq. AK and AG
= sq. ME + sq. FP .

sq. upon AB = sq.
upon BH + sq. upon AH.

= a + D .

a. Devised by author,
March 26, 1926, 10 p.m.

Fig. 298

= trl. TON) + (sq. SQ =
sq. upon AB =

In fig. 298, the sq.
on AH is translated to posi-
tion of GC, and the sq. on HB
to position of GD. Complete
the figure and conceive the
sum of the two sq>s EL and GC
as the two reefs EM + TC
+ sq. LN and the dissection
as numbered.

Sq. AK = (tri. AGP
= tri. DTM) + (tri. CKQ
= trl. TDE) 4- (tri. KBR
= trl. CTO) -f (trl. BAS
sq. LH) = sq. EL + sq. GC .
sq. upon BH + sq. upon AH.

h^ = a^ + b^.

a. Devised by author, March 22, 1926.

b. As sq. EL, having a vertex and a side in
common with a vertex and a side of sq. GC, either ex-
ternally (as in fig. 298), or Internally, may have 12
different positions, and as sq. GC may have a vertex

212

THE PYTHAGOREAN PROPOSITION

and a side in common with the fixed sq. AK, or In
common with the given triangle ABH, giving 15 differ-
ent positions, there is possible l80 - 3 = 177 dif-
ferent figures, hence 176 proofs other than the one
given above, using the dissection as used here, and
178 more proofs by using the dissection as given in
proof Ten , fig. 111.

c. This proof is a variation of that given
in proof Eleven , fig. 112.

I

Ifi-T-ilo

4l.( iP?

Fig. 299

In fig, 299, the
construction is evident,
as PO is the translation
of the sq. on AH, and KE
is the translation of the
sq. on BH.

Since rect. ON
= rect, QE, we have sq.

AK = (tri. LK7 = trl. CPL)
+ (trl, KBW = trl. LPC)

+ (trl. BAT = trl. KQR)

+ (trl, ALU = trl, RSK)

+ (sq. TV = sq. MO)

= rect. KR + rect. PP
+ sq. MO = sq. KE + sq.

PO.

sq. upon AB = sq. upon BH + sq. upon AH.
h^ = a^ + b^.

a. Devised by the author, March 27, 1926.

In fig. 300 the translation and construction
are easily seen,

Sq. AK = (tri. CKN = trl. LPG) + (trap. OTUM
= trap. RESA) + (trl. VOB = trl. RAD + (quad. ACNV
+ trl. TKU = quad. MKPL) = sq. DS + sq. MP.

GEOMETRIC PROOFS

213

Fig. 300

sq. upon AB = sq.
upon HB + sq. upon HA. h®
= 4- b^.

a. Devised by the
author, March 27, 1926,
10:40 p.m.

Fig. 301

AR = AH and AD = BH. Com-
plete sq*s on AR and AD. Extend DE
to S and drav SA and TR.

Sq. AK = (tri. QPB = trl.

VDR of sq. AP) + (trap. ALPQ = trap.
ETAU of sq. AE) + (trl. CMA = trl.
SGA of sq. AP) + (trl. CNM = trl.

UAD of sq. AE) + (trap. NKOL = trap.
VRPS of sq. AP) + (trl. 0KB = trl.
DSA of sq. AP) = (parts 2 4- 4 = sq.
AE) 4- (parts 14-3 + 54-6 = sq. AP).

sq. upon AB = sq. upon HB
+ sq. upon HA. /. h^ = a^ + b^.
Q.E.D.

. a. Devised, by author, Nov. l6, 1933.

Iwg_Hu.n4rei_Fau.r

Fig. 302

In fig. 302, complete
the sq. on EH, draw BD par. to
AH, and draw AL and KP perp.
to DB.

Sq. AK = sq. HG - (4
trl. ABH = 2 rect. HL) = sq.

EL + sq. DK + 2 rect. PM - 2
rect. HL = sq. EL + sq. DK.

/. sq. upon AB = sq.
upon HB + sq. upon HA. h^

= a^ + b^.

214

THE PYTHAGOREAN PROPOSITION

a. See Edwards* Geom. , 1895 ^ p. 158, fig.

(19).

b. By changing position of sq. PG, many other
proofs might be obtained.

c. This is a variation of proof, fig. 240.

In fig. 505, let W and X be sq*s with sides
equal resp*y to AH and BH. Place them as in figure,

A being center of sq. W, and 0, middle of AB as cen-
ter of PS. ST = BH, TP = AH. Sides of sq’s PV and
QS are perp. to sides AH and BH.

It is obvious that:

Sq. AK = (parts l+2+5+4= sq. PV ) + sq.
QS = sq. X + sq. W.

sq. upon AB = sq. upon HB + sq. upon HA.

. . h = a + b .

a. See Messenger of Math., Vol. 2, p. 105,
1873, and there credited to Henry Perlgal, P.R.S.A.S.

Iwa.H«.nired.Six
Case (6), (b).

In fig. 304, the construction
is evident. Sq. AK = (trl. ABH
= trap. KEMN + trl. KOP) + (trl. BOH
= trl. KLN) + quad. GOKC common to
sq*s AK and CP + (trl. CAG = trl. CKE)
= sq. MK + sq. CP.

Fig. 304

GHEOMETRIC PROOFS

215

sq. upon AB = sq, upon BH + sq. upon AH.

/. + b^. Q.E.D.

a. See Hopkins’ Plane Geom. , I 89 I, p. 92, fig
fig. VIII.

b. By drawing a line EH, a proof through par-
allelogram, may be obtained. Also an algebraic proof.

c. Also any one of the other three triangles,
as GAG may be called the given triangle, from which
other proofs would follow. Furthermore since the trl.
ABH may have seven other positions leaving side of
sq. AK as hypotenuse, and the sq. MK may have 12 po-
sitions having a side and a vertex in common with sq.
CP, we would have 8^ proofs, some of which have been
or will be given; etc., etc., as to sq. CP, one of
which is the next proof.

In fig. 505 , through H
draw LM, and draw CN par. to BH
and KO par. to AH.

Sq. AK = rect. KM + rect.
CM = paral. KH + paral . CH = HB
X KO + AH X CN = sq. on BH + sq.
on AH = sq. MD + sq. MG.

/. sq. upon AB = sq. upon
BH + sq. upon AH. /. h^ = a^ + b^.

a. Original with the
author January 31, 1926, 3 p.m.

Case ( 7 ), (a).

In fig. 306 , extend AB to X, draw NTJ and KS
each = to AH and par. to AB, CV and HT perp. to AB,
GR and PP par. to AB, and LW and AM perp. to AB.

2l6

THE PYTHAGOREAN PROPOSITION

Sq. WK = (tri. OKS
= tri. PPL = trap. BYDX
of sq. BD + tri. FON of
sq. GP) + (tri. TlOi = tri.
GRA. = tri. BEX of sq. BD
+ trap. WQRA of sq. GP)

+ (tri. WUH = tri. LWG of
sq. GP) +ftri. WCV = tri.
WLN of sq. GP)+(sq. VT
= paral. RO of sq. GP)

= sq. BD + sq. GP.

/. sq. upon AB = sq.
upon HB + sq. upon HA.

a. Original with the author, Aug. 8 , 19OO.

b. As in fig. 305 many other arrangements are
possible each of which will furnish a proof or proofs.

J

{A) — Proofs determined by ariuments based upon a
square.

This type Includes all proofs derived from
figures in which one or more of the squares are not
graphically represented. There are two leading class-
es or sub-types in this type — first, the class in
whjtch the determination of the proof is based upon a
square; second, the class in which the determination
of the proof is based upon a triangle.

As in the I-type, so here, by inspection we
find 6 sub-classes in our first sub-type which may be
symbolized thus;

(1) The h-square omitted, with

(a) The a- and b-squares const ’d outwardly —

3 cases.

(b) The a-sq. const *d out’ly and the b-sq.
overlapping — 3 cases.

(c) The b-sq. const *d out*ly and the a-sq.
overlapping — 3 cases.

(d) The a- and b-squares overlapping — 3 cases.

GEOMETRIC PROOFS

217

(2) The a-sq. omitted, vlth

(a) The h- and b-sq*s const *d out*ly--5 cases.

(b) The h-sq. const *d out*ly and the b-sq.
overlapping — 5 cases.

(c ) The b-sq. const *d out*ly and the h-sq.
overlapping — 3 cases.

(d) The h- and b-sq*s const ’d and overlapping
--3 cases.

(3) The b-sq. omitted, with

(a) The h- and a-sq*s const *d out'ly--3 cases.

(b ) The h-sq. const *d out’ly and the a-sq.
overlapping — 3 cases.

(c ) The a-sq. const *d out * ly and the h-sq.
overlapping- -3 cases.

(d) The h- and a-sq*s const *d overlapplng--
3 cases.

(4) The h- and a-sq*s omitted, with

(a) The b-sq. const *d out*ly.

(b) The b-sq. const *d overlapping.

(c ) The b-sq. translated — In all 3 cases.

(5) The h- and b-sq*d omitted, with

(a) The a-sq. const *d out’ly,

(b) The a-sq. const *d overlapping.

(c ) The a-sq. translated — In all 3 cases.

(6) The a- and b-sq's omitted, with

(a) The h-sq, const *d out*ly.

(b) The h-sq. const *d overlapping.

(c ) The h-sq. translated — In all 3 cases.

The total of these enumerated cases Is 45. We
shall give but a few of these 45, leaving the re-
mainder to the Ingenuity of the Interested student.

(7) All three squares omitted.

218

THE PYTHAGOREAN PROPOSITION

Case (l), (a).

In fig. 307, produce
GF to N a pt., on the perp. to
AB at B, and extend DE to L,
draw HL and AM perp. to AB.

The tri ' s AMG and ABH are
equal .

Sq. HD + sq. GH
= (paral . HO = paral . LP )

+ paral. MN = paral. MP = AM
xAB=AB xAB= (AB)^.

sq. upon AB = sq.

upon BH + sq. upon AH. h^ = a^ + h^.

a. Devised by author for case (l), (a),

March 20, 1926.

b. See proof No. 88, fig. I88. By omitting
lines OK and HN in said figure we have fig. 307.
Therefore proof No. 209 is only a variation of proof
No. 88, fig. 188.

Analysis of proofs given will show that many
supposedly new proofs are only modifications of some
more fundamental proof.

Iwfi.HU.ELiL£Si-l£[l

(Not a Pythagorean Proof . )

While case (l), (b) may be proved in some

other way, we have selected the following as being
quite unique. It is due to the ingenuity of Mr.
Arthur R. Colburn of Washington, D.C., and is No. 97
of his 108 proofs.

It rests upon the following Theorem on Paral-
lelogram, which is: ”lf from one end of the side of
a parallelogram a straight line be drawn to any point
in the opposite side, or the opposite side extended,
and a line from the other end of said first side be
drawn perpendicular to the first line, or its

GEOMETRIC PROOFS

219

extension, the product of these two drawn lines will
measure the area of the parallelogram." Mr. Colburn
formulated this theorem and its use is discussed in
Vol. 4, p, 45, of the "Mathematics Teacher," Dec.,
1911. I have not seen his proof, but have demonstrat-
ed it as follows:

In the paral.
ABCD, from the end A of
the side AB, draw AP to
side DC produced, and
from B, the other end
of side AB, draw BG
perp. to AP. Then AF
X BG = area of paral.
ABCD.

Proof: Prom D lay off DE = CP, and draw AE
and BP forming the paral. ABPE = paral. ABCD. ABP
is a triangle and is one-half of ABPE, The area of
tri. PAB = JPA ^ BG; therefore the area of paral.
ABPE = 2 times the area of the tri, PAB, or PA x bG.
But the area of paral. ABPE = area of paral, ABCD.

AP X BG measures the area of paral. ABCD.

Q.E.D.

By means of this Parallelogram Theorem the
Pythagorean Theorem can be proved in many cases, of
which here is one.

Case (1), (b).

In fig. 509, extend GP
and ED to L completing the paral.
AL, draw PE and extend AB to M.
Then by the paral. theorem:

(1) EP X AM = AE X AG.

(2) EP X bM = PL X BP.

(1) - (2) = (5) EP{AM - BM)

= AE X AG - PL X BP

Fig. 509

220

THE POTHAGOREAN PROPOSITION

(5) = (4) (EP = AB) X aB = AGPH 4 - BDEH, or sq. AB
= sq. HG + sq. HD.

sq, upon AB = sq. upon BH + sq. upon AH.

a. This is No. 97 of A. R. Colburn’s 108

proofs.

b. By inspecting this figure we discover in
It the five dissected parts as set forth by my Law
of Dissection. See proof Ten , fig. 111.

lM-.tty.ndrg.d_Iw^lv^

Fig. 310

Case (2), (b ) .

Trl. HAC = tri. ACH.

Trl. HAC = i sq. HG.

Trl. ACH = J rect. AL.

rect. AL = sq. HG. Similarly
rect. BL = sq. on HB. But rect. AL
+ rect, BL = sq, AK.

sq. upon AK = sq. upon HB
+ sq. upon HA. h^ = a^ + b^.

Q.E ,D ,

a. Sent to me by J. Adams from The Hague,
Holland. But the author not given. Received it
March 2, 1934.

Iwa.Hu.ndrg.i_Ihirtg.g.Q.

Case (2), (c).

In fig. 311 1 produce GA
to M making AM = HB, draw BM,
and draw KL par. to AH and CO
par. to BH.

Sq. AK = 4 tri. ABH + sq.

NH = 4 X ^ ^ + (AH - BH)®

= 2AH X bH + AH® -
BH® = BH® + AH®.

2AH X bH +

GEOMETRIC PROOFS

221

sq. upon AB = sq. upon BH 4- sq. upon AH.

A

a. Original with author, March, 1926.

b. See Sci. Am. Sup., Vol. 70, p. ^83, Dec.
10, I9IP, fig. 17, in which Mr. Colburn makes use of
the tri. BAM.

c. Another proof, by author, is obtained by
comparison and substitution of dissected parts as
numbered.

t£|^n

Case (4), (b).

In fig. 312, produce PG to P
making GP = BH, draw AP and BP.

Sq. GH = b^ = tri. BHA + quad.
ABPG = tri. APG + quad. ABPG = tri.
APB + tri. PPB = ic^ + i(b + a)(b-a).
A b^ = + l-b^ - .% c^ = a^

+ b^.

Fig. 312 A sq. upon AB = sq. upon HB

+ sq. upon HA.

a. Proof 4, on p. 104, in "A Companion of
Elementary School Mathematics,” (1924) by P. C, Boon,
B.A., Pub, by Longmans, Green and Co.

Iwa_H!i!ldred_Fj_f teen

Fig. 313

(AH® + HB®)
= BH® + AH®.

In fig. 313# produce HB to P
and complete the sq. AP, Draw GL
perp. to AB, PM par. to AB and NH
perp. to AB.

Sq. AF = AH® = 4

AO X HO

+ [LO® = (AO - HO)®] = 2A0 x HO + AO®
- 2A0 X HO + HO® = AO® + HO® = (AO
= AH® -f AB)® + (HO = AH X HB 4 - AB)®

= AH^ * AB® + AH® X hB® 4 - AB® = AH®

AB‘

1 = (AH® 4 - BH®) 4 - AB*

AB‘

222

THE PYTHAGOREAN PROPOSITION

sq, upon AB = sq. upon HB + sq. upon HA.
h® = + b®.. Q.E.D.

a. See Am. Math. Mo., Vol. VI, 1899, P. 69 ,
proof CIII; Dr. Leltzmann, p. 22, fig. 26,

b. The reader will observe that this proof
proves too much, as It first proves that AH^ = AO^

+ HO^, which is the truth sought. Triangles ABH and
AOH are similar, and what is true as to the relations
of the sides of tri. AHO must be true, by the law of
similarity, as to the relations of the sides of the
tri. ABH.

IbLfi -H y. Q. d i Lx. t®. ® Q.

Case (6), (a). This is a popular

figure with authors.

In fig. 31 ^ > draw CD and KD
par. respectively to AH and BH, draw
AD and BD, and draw AP perp. to CD
and BE perp. to KD extended.

Sq. AK = 2 tri. CDA + 2 tri.
SDK =CDxAP + Kr>>'EB = CD® + KD®.

sq. upon AB = sq. upon BH
+ sq. upon AH. h^ = a^ + b^.

a. Original with the author,
August 4, 1900 .

Iwfi.Hu.!ldred_S§vent££a

In fig. 315 ^ extend AH and
BH to E and P respectively making HE
= HB and HP = HA, and through H draw
LN perp. to AB, draw CM and KM par.
respectively to AH and BH, complete
the rect. PE and draw LA, LB, HC and
HK.

Sq. AK = rect. BN + rect. AN
= paral. BM + paral. AM = (2 tri. HMK
= 2 tri. LHB = sq. BH) + (2 tri. HAL
= 2 tri. LAH = sq. AH).

Fig. 315

GEOMETRIC PROOFS

223

/. sq. upon AB = sq. upon BH 4- sq. upon AH.
• • n — cL “T D •

a. Original with author March 26 , 1926,

9 P-m.

Fig. 316

In fig. 316 , complete the sq's HF and AK; in
fig. 317 complete the 3q*s HF, AD and CG, and draw
HC and DK. Sq. HF - 4 tri. ABH = sq, AK = h^. Again
sq. HF - 4 tri. ABH = a^ + h^. .•. h^ = a^ +

.*. sq. upon AB = sq. upon BH + sq. upon AH.

a. See Math. Mo., I 858 , Dem. 9> Vol. I, p.
159^ and credited to Rev. A. D. Wheeler of Brunswick,
Me., in work of Henry Bead, London, 1735.

b. An algebraic proof: a^ + b^ + 2ab = h^

+ 2ab. h^ = a^ + b^,

c. Also, two equal squares of paper and scis-
sors .

Jwa.Haadrad.Iiiaaiaen

In fig. 318 , extend HB to N and complete the

sq. HM.

Sq. AK = sq. HM - 4 — ■ ^ = (LA + AH)®

- 2HB X HA = LA® + 2LA x aH + AH® - 2HB x hA = BH®

+ AH®.

224

THE PYTHAGOREAN PROPOSITION

1895 ,

159 , fig.

/. sq. upon AB = sq.
upon BH + sq, upon AH.

a. Credited to T. P.
Stowell, of Rochester, N.Y.

See The Math. Magazine, Vol. I,
^ I I \ 1882, p. 58; 01 ney*s Geom. ,

\ I t \ Part III, 1872, p. 251, 7 th

✓ iv ^©"thod; Jour, of Ed*n, Vol.

^ ' XXVI, 1877, P. 21 , fig. IX;

^ C K also Vol. XXVII, 1888, p. 327,

Mx 18th proof, by R. E. Blnford,

Independence, Texas; The
Fig. 318 School Visitor, Vol. IX, I888,

p. 5 , proof II; Edwards* Geom.,
1895, P. 159, fig. (27); Am. Math. Mo., Vol. VI,

1899, p. 70, proof XCIV; Heath* s Math. Monographs,

No. 1 , 1900, p. 23, proof VIII; Scl. Am. Sup., Vol.
70, p. 359 , fig. 19IO; Henry Boad*s work, London,

1755.

b. For algebraic solutions, see p. 2 , In a
pamphlet by Artemus Martin of Washington, D.C., Aug.
1912, entitled "On Rational Right-Angled Triangles";
and a solution by A. R. Colburn, In Scl. Am. Supple-
ment, Vol. 70, p. 359 , Dec. 3 , 1910.

c. By drawing the line AK, and considering

the part of the figure to the
right of said line AK, we have
figure from which the

\ proof known as Garfield* s So-

\ \ lutlon follows — see proof Two

\ ^ Hundred Thirty-One ^ fig. 330.

\

, f l\ Iwa.HiLalred^Iw^nti

f I i \

I I V In fig. 319, extend

j ) HA to L and complete the sq.

^

Ca Sq. AK = aq. LN

Sq. AK = sq. LN
_ 4 X = (HB + HA) =

- 2 HB X HA = HB® + 2 HB x hA

Elg. 319

GEOMETRIC PROOFS

225

+ - 2HB X HA = sq. HB + sq. HA. /. sq. upon AB

= sq. upon BH + sq. upon AH. + b^.

a. See Jury Wlpper, l880, p. 35, fig. 32, as
given in "Hubert’s Rudlmenta Algebrae," Wurceb, 1762;
Versluys, p. 70, fig. 75.

b. This fig, 319 is but a variation of fig.
240, as also is the proof.

Case (6), (b).

In fig. 320, complete the
sq. AK overlapping the tri. ABH,
draw through H the line LM perp. to
AB, extend BH to N making BN = AH,
and draw KN perp. to BN, and CO
perp. to AH. Then, by the paral-
lelogram theorem. Case (l), (b),
fig. 308, sq. AK = paral. KM
+ paral. CM = (BH x kn = a^) + (AH x cO = b^ ) = a^

+ b^.

.*. sq. upon AB = sq. upon BH + sq. upon AH.

a. See Math. Teacher, Vol. 4, p. 45, 191I,
where the proof is credited to Arthur H. Colburn.

b. See fig. 324; which is more fundamental,
proof No. 221 or proof No. 225?

c. See fig. 114 and fig. 328.

Ivifi.Hy.Qclred.Iw£nt]t-Twg

In fig. 521, draw CL perp.
to AH, produce BH to N making BN
= CL, and draw KN and CH. Since CL
= AH and KN = BH, then J sq. BC
= tri. KBH + tri. AHC = J BH^ + JAH^,
or ih^ = ia^ + ib^. h^ = a^ + b^.

.*. sq. upon AB = sq. upon HB
+ sq. upon HA.

a. Proof 5 , on p. 104, in

Fig. 521

226

THE PYTHAGOREAN PROPOSITION

”A Companion to Elementary Mathematics" (1924) hy
F. C, Boon, A.B., and credited to the late P. C. Jack-
son ("Slide Rule Jackson").

Iwa-HyLQ.iced_Twe.nty.-Thr eg.

*^k =

In fig. 322, dnaw CL and KL
par. to AH and BH respectively, and
through H draw LM.

Sq.. AK = rect, KM + rect. CM
= paral. KH + paral. CH = BH x nl
+ AH NH = BH® + AH®,

sq. upon AB = sq. upon BH
4- sq. upon AH. h^ = + h^.

I * \ * h + Id .

a. This is known as Haynes*
Fig. 522 Solution. See the Math. Magazine,

Vol. I, p, 6o, 1882; also said to
have been discovered In l877 Geo. M. Phillips, Ph.
Ph.D., Prin. of the West Chester State Normal School,
Pa.; see Heath* s Math. Monographs, No. 2, p. 58,
proof XXVI; Fourrey, p. 76.

h. An algebraic proof is easily obtained.

Iwfi-HlllliLii-IwenlYrtftii!!

Fig. ’"325

In fig. 523, construct sq.
AK. Extend AH to G making HG = HB;
on CK const, rt. tri. CKL (= ABH)
and draw the perp. LHM, and extend
LK to G.

Now LG = HA, and it is obvi-
ous that: sq. AK (= h^ ) = rect. MK
+ rect. MC = paral. HK + paral. HC
= HB X hG + HA X CL = b^ + a^, or
h® = a® + b®. Q.E.D.

sq, upon AB = sq. upon HB
+ sq. upon HA,

a. This fig, (and proof) was devised by Gus-
tav Cass, a pupil in the Junior-Senior High School,

GEOMETRIC PROOFS

227

South Bend, Ind., and sent to the author, by his
teacher, Wilson Thornton, May l6, 1959.

Case (6), (c).

For convenience designate
the upper part of fig. 524, l.e.,
the sq. AK, as fig. 524a, and the
lower part as 324b.

In fig. 524a, the con-
struction Is evident, for 324b Is
made from the dissected parts of
324a. GH* Is a sq. each side of
which = AH, LB* Is a sq. each
side of which = BH.

Sq. AK = 2 trl. ABH + 2
trl. ABH + sq. MH = rect. B*N
+ rect. OF + sq. LM = sq. B*L
+ sq. A*F.

Fig. 324t A sq. upon AB = sq. upon

BH + sq. upon AH, h^ = a^ + b^.

a. See Hopkins* Plane Geom. , 1891# p. 91^
fig. V; Am. Math. Mo., Vol. VI, 1899, p. 69 , XCI;

Beman and Smith* s New Plane Geom., 1899, P. 104, fig.
3; Heath* s Math. Monographs, No. 1, 1900, p. 20,
proof IV. Also Mr. Bodo M. DeBeck, of Cincinnati,

0., about 1905 without knowledge of any previous so-
lution discovered above form of figure and devised a
proof from It. Also Versluys, p. 31, fig* 29; and
"Curiosities of Gteometrlques, Fourrey, p. 83, fig. b,
and p. 84, fig. d, by Sanvens, 1755.

b. History relates that the Hindu Mathema-
tician Bhaskara, born 1114 A.D,, discovered the above
proof and followed the figure with the single word
"Behold," not condescending to give other than the
figure and this one word for proof. And history
furthermore declares that the Geometers of Hindustan
knew the truth and proof of this theorem centuries
before the time of Pythagoras — may he not have learned
about It while studying Indian lore at Babylon?

ft

Pig. 324a

X I 1

IF

X 1

228

THE PYTHAGOREAN PROPOSITION

Whether he gave fig. 324b as well as fig.

324a, as I am of the opinion he did, many late authors
think not; with the two figures, 324a and 324b, side
by side, the word "Beholdl” may be justified, espe-
cially when we recall that the tendency of that age
was to keep secret the discovery of truth for certain
purposes and from certain classes; but with the fig.
324b omitted, the act Is hardly defensible — not any
more so than "See?" would be after fig. 318.

Again, authors who give 324a and "Behold!"
fall to tell their readers whether Bhaskara’s proof
was geometric or algebraic. Why this silence on so
essential a point? For, If algebraic, the fig. 324a
Is enough as the next two proofs show. I now quote
from Beman and Smith: "The Inside square Is evident-
ly (b - a)^, and each of the four triangles Is ^ab;

/. h^ - 4 X ^ab = (b - a)^, whence h^ = a^ + b^."

It Is conjectured that Pythagoras had discov-
ered It Independently, as also did Wallis, an English
Mathematician, In the 17th century, and so reported;
also Miss Coolldge, the blind girl, a few years ago:
see proof Thirty-Two , fig. 135.

Iwft^!!y.nirsd.Iw£nt3i-Six

In fig. 325, It Is ob-
vious that trl*s 7 + 8 = rect.
GL. Then It Is easily seen,
from congruent parts, that:
sq. upon AB = sq. upon BH + sq.
upon AH. h^ = a^ + b^.

a. Devised by R. A.
Bell, Cleveland, 0., July 4,
1918. He submitted three more
of seune type.

In fig. 326, PG» = PH* = AB = h, DG* = EP
= PN = OH* = BH = a, and DM* = EH* = G*N = PO = AH
= b.

GEOMETRIC PROOFS

229

H'

£^'-7

Pig. 326

Then are trl ' s
PGD, G'PN = PH'E and
H'PO each equal to PG'D
= trl, ABH. ■

Now 4 trl. PG'D
+ sq. Q'H' = sq. EN
+ sq. DO = a^ + b®.

But 4 trl. PG'D + sq.
G'H' = 4 trl. ABH + sq.
GH = sq. AK = h®. .-. h®

= a® + b®. sq. upon AB = sq. upon BH + sq. upon

AH.

a. Devised by author, Jan. 5, 1934.

b. See Versluys, p. 69 , fig. 73.

Draw AL and BL par.
resp’ly to BH and AH, and com-
plete the sq. LN. ABKC =

= (b + a)® - 2ab; but ABKC
= (b - a)® + 2ab.

.-. 2h® = 2a® + 2b®,
or h® = a® + b®. .*. sq. upon

AB = sq. upon BH + sq. upon
AH.

a. See Versluys, p.
72 , fig. 78 , attributed to
Saimderson (1682-1739 )y arid
came probably from the Hindu
Mathematician Bhaskara.

In fig. 328 , draw CN par. to
BH, KM par. to AH, and extend BH to

L.

Sq. AK

4

HB X HA
2

+ sq. MH

= 2HB X ha + (AH - BH)® = 2HB x HA
+ HA® - 2HB X ha + HB® = HB® + HA®.

Pig. 328

230

THE PYTHAGOREAN PROPOSITION

/. sq. ■upon AB = sq. upon BH + sq, upon AH.

a. See Olney^s Geom. , Part III, 1872, p. 250,
1st method; Jour, of Ed*n, Vol. XXV, 1887, p. ^04,
fig. rv, and also fig. VI; Jour, of Ed*n, Vol. XXVII,
1888, p. 327, 20th proof, by R. E. Blnford, of Inde-
pendence, Texas; Edwards* Geom., 1895, P* 155, fig*
(3); Am. Math. Mo., Vol. VI, l899, P. 69, proof XCII;
Scl. Am. Sup., Vol. 70, p. 359, Dec. 3, 1910, fig. 1;
Versluys, p. 68, fig. 72; Dr. Leltzmann*s work, 1930,
p. 22, fig. 26; Pourrey, p. 22, fig. a, as given by
Bhaskara 12th century A.D. In VI ja Ganlta. For an
algebraic proof see fig. 32, proof No. 3^, under Al-
gebraic Proofs.

b. A study of the many proofs by Arthur R.
Colburn, LL.M., of Dlst. of Coliombla Bar, establishes
the thesis, so often reiterated In this work, that
figures may take any form and position so long as
they Include triangles whose sides bear a rational
algebraic relation to the sides of the given triangle,
or whose dissected areas are so related, through
equivalency that h^ = a^ + b^ results.

(B) — Proofs based uvon a triangle throuih the calcu-
lations and comparisons of equivalent areas.

Draw HC perp. to AB. The
three trl*s ABH, BHC and HAG are sim-
ilar.

We have three slm. trl's
erected upon the three sides of trl.
ABH whose hypotenuses are the three
sides of trl. ABH.

Now since the area of trl. CBH + area of trl.
CHA = area of trl. ABH, and since the areas of three
slm. trl*s are to each other as the squares of their
corresponding sides, (in this case the three hypote-
nuses), therefore the area of each trl. Is to the sq.
of Its hypotenuse as the areas of the other two trl*s
are to the sq*s of their hypotenuses.

GEOMETRIC PROOFS

231

Now each sq. is = to the trl. on whose hypote-
nuse it is erected taken a certain number of times,
this number being the same for all three. Therefore
since the hypotenuses on which these sq*s are erect-
ed are the sides of the tri. ABH, and since the sum

of tri *3 erected on the legs is = to the tri. erected

on the hypotenuse. the Siam of the sq's erected on
the legs = the sq. erected on the hypotenuse, h^

= a^ + b^. Q.E.D.

a. Original, by Stanley Jashemski, age 19, of
Youngstown, 0., June 4, 1934, a young man of superior
intellect .

b. If m + n = p and m : n : p = a^ : b^ : h ,

then m + n : a^ + b^ = n : b^ = p : h^.

m + n a^ + b^ , a^ + b^ . ^2

= , or 1 = T . A h

P h^ h^

= a^ + b^. This algebraic proof given by E. S. Loom-
is,

In fig. 330, extend HB to
D making BD = AH, through D draw
DC par. to AH and equal to BH, and
draw CB and CA.

Area of trap. CDHA = area
of ACB + 2 area of ABH.

i (AH + CD )HD = Jab® + 2
X JAH X HB or (AH + HB)® = AB®

+ 2AH X hB, whence AB® = BH® + AH®.

sq. upon AB = sq. upon
BH + sq. upon AH. h® = a® + b®.

a. This Is the "Garfield
Demonstration," — hit upon by the General In a mathe-
matical discussion with other M.C.'s about I876. See
Jour, of Ed'n, Vol. Ill, I876, p. I6I; The Math. Mag-
azine, Vol. I, 1882, p. 7; The School Visitor, Vol.

IX, 1888, p. 5, proof III; Hopkins' Plane Geom. , 1893,
p. 91, fig. VII; Edwards' Geom., 1895, P- 156, fig.
(11); Heath's Math. Monographs, No. 1, I9OO, p. 25,

232

THE PYTHAGOREAN PROPOSITION

proof X; Pourrey, p. 95; School Visitor, Vol. 20, p.
167 ; Dr. Leltzmann, p. 23, fig. 28a, and also fig.

28b for a variation.

b. For extension of any triangle, see V. Jel-
Inek, Casopls, 28 (1899 ) 79—; Pschr. Math. (1899)
456 .

c. See No. 219, fig. 3l8.

a. See

Kruger, 17^6.

By geometry, (see Wentworth* s
Revised Ed*n, 1895, p. 161, Prop'n
XIX), we have AH^ + HB^ = 2HM^ + 2AM^.
But In a rt. tri. HM = AM. So b^

+ a^ = 2AM^ + 2AM^ = 4AM^ = 4(^)^

= AB^ = h^. h^ = a^ + b^.

Versluys, p. 89 , fig. 100, as given by

Iwft^tliLaired^Ihirtx-Ihrag

Given rt. trl. ABH. Extend
BH to A* making HA* = HA. Drop A*D
perp. to AB Intersecting AH at C.
Draw AA* and CB.

Since angle ACD = angle HCA*,
then angle CA*H = angle BAH, There-
fore trl*s CHA* and BHA are equal.
Therefore HO = HB.

Fig. 332

Quad. ACBA* = (trl. CAA*
= trl, CAB) + trl. BHC + CHA* =

+ ^ h(AD 4- BD)

2 2

h-^ = a-^ + b^.

a. See Dr. W. Leltzmann* s work, p. 23, fig.
27, 1930, 3i*ci edition, credited to C. Hawkins, of
Eng., who discovered It In I909.

b. See Its algebraic proof Fifty , fig. 48.
The above proof Is truly algebraic throu^ equal
areas. The author.

GEOMETRIC PROOFS

253

Let C, D and E be the cen-
ters of the sq*s on AB, BH and HA.
Then angle BHD = 45°, also angle
EHA. line ED through H Is a st.
line. Since angle AHB = angle BOA
the quad, is Inscrlptlble in a cir-
cle vhose center is the middle pt.
of AB, the angle CHB = angle BHD
= 45^. A CH is par. to BD. a an-
gle CHD = angle HDB = 90°. Draw
AG and BP perp. to CH. Since tri*s AGC and CPB are
congruent, CG = FB = DB and HG = AG = AE, then CH
= EA + BD.

Now area of ACBH = ^(AG + PB) = ^ x eD

= area of ABDE. Prom each take away tri. ABH, we get
trl. ACB = tri. BHD 4- tri. HEA. 4 times this eq*n
gives sq. upon AB = sq. upon HB + sq. upon HA. h^

= a^ + b^.

a. See Fourrey, p. 78, as given by M. Plton-
Bressant; Versluys, p. 90, fig. 105, taken from Van
Plton-Bressant, per Fourrey, I907,

b. See algebraic prqof No. 67, fig. 66.

Iwa.tliLQdrei.Ihirt^-Fiyfe

Pig. 333 and 554 are same
in outline. Draw HP perp. to AB,
and draw DC, DP and PC. As in
proof, fig. 533, HC is a st. line
par. to BD. Then trl. BDH = trl.
BDC. (1) As quad. HPBD is In-

scrlptlble in a circle whose center
is the center of HB, then angle BPD
= angle DPH = 45° = angle PBC. PD
Is par. to CB, whence trl. BCD

. —( 2 ).

tri. BCP

25 ^

THE PYTHAGOREAN PROPOSITION

tri. BOP = tri. BDH. In like manner tri.
AGP = tri. AHE. /. tri. ACB = tri. BDH + tri. HEA.

(5), 4 X (3) gives sq. upon AB = sq. upon HB

+ sq. upon HA. + b^.

a. See Pourrey, p. 79> as given by M. Plton-
Bressant of Vltteneuve-Salnt-Georges; also Versluys,
p. 91, fig. 104.

b. See algebraic proof No. 66, fig. 67.

Iwfi_Hu.ndr^d_Ih i.r Si.x

I \ \

I \ ' U

Fig. 555

In fig. 535, extend BH to P
making HP = AH, erect AG perp. to AB
making AG = AB, draw GE par. to HB
and GD par. to AB. Since tri ' s ABH
and GDP are similar, GD = h (l - a/b),
and PD = a (1 - a/b ) .

Area of fig. ABPG = area ABH
+ area AHPG = area ABDG +

^ab + |•b[b + (b - a)] =

(1 - a/b)] + ia(b - a) (l

GDP.
+ h
a/b).

area

ih[h

+ b"

sq. upon AB = sq.

(1 ) . Whence

upon BH + sq. upon AH.

a. This proof Is due to J. G. Thompson, of
Winchester, N.H.; see Jour, of Ed*n, Vol. XXVIII,
1888, p. 17, 28th proof; Heath* s Math. Monographs,
No. 2, p. 54, proof XXIII; Versluys, p. 78, fig. 87,

by Rupert, I9OO.

b. As there are possible several figures of
above type. In each of which there will result two
similar triangles, there are possible many different
proofs, differing only In shape of figure. The next
proof Is one from the many.

IwO-.!iy.ndred_Th l_r ty^-S§.ven

In fig. 336 produce HB to P making HP = HA,
through A draw AC perp. to AB making AC = AB, draw
CP, AG par. to HB, BE par. to AH, and BD perp. to AB.

GEOMETRIC PROOFS

255

Since trl*3 ABH and BDP are similar,
we find that DP = a (l - a/h ) and BD
= h(l - a/b).

Area of trap. CPHA = 2 area
ABH + area trap. AGFB = area ABH.

+ area trap. ACDB + area BDP.

Whence area ACG + area AGPB
= area ACDB + area BDP or l-ab
+ ib[h + (b - a)] = ih[h + h (l - a/b)]

+ ia (b - a ) (l - a/b ) .

This equation is equation (l ) in the preced-
ing solution, as it ought to be, since, if we draw
BE par. to AH and consider only the figure below the
line AB, calling the tri. ACG the given triangle, we
have identically fig. 555, above.

/. sq. upon AB = sq. upon BH + sq. upon AH.

A h^ = a^ + b^.

a. Original with the author, August, 1900.
See also Jour, of Ed*n, Vol. XXVIII, 1888, p. 17,
28th proof.

In fig. 557, extend
HB to N making HN = AB, draw
KN, KH and BG, extend GA to M
and draw BL par. to AH. Tri.
KBA + tri. ABH = quad. BHAK
= (tri. HAK = tri. GAB)

+ (tri. DGB = tri. HKB )

= quad. ABDG = tri. HBD + tri.
GAH + tri. ABH, whence tri.
BAK = tri. HBD 4- tri. GAH.

.*. sq. upon AB = sq.
upon BH -f- sq. upon AH. h^

= a^ + b^.

a. See Jury Wipper,
l880, p. 55, fig* 50, as found in the works of Joh.

J. I. Hoffmann, Mayence, 1821; Pourrey, p. 75*

I'k'/'

Fig. 557

256

THE PYTHAGOREAN PROPOSITION

In fig. 558> construct the
three equilateral triangles upon the
three sides of the given triangle
ABH, and draw EB and FH, draw EG
perp. to AH, and draw GB.

Since EG and HB are parallel,
trl. EBH = trl. BEG = ^ trl, ABH.

.’. trl. GBH = trl. HEG.

(l) Trl. HAP = trl. EAB
= trl. EAK + (trl. BGA = i trl. ABH)
+ (trl. BKG = trl. EKH) = trl. EAH
+ i trl. ABH.

(2) In like manner, trl. BHP = trl. DHB + i
trl. ABH. (1) + (2) = (5) (trl. HAP + trl. BHP = trl.
BAP + trl. ABH) = trl. EAH + trl. DHB + trl. ABH,
whence trl. FBA = trl. EAH + trl, DHB.

But since areas of similar surfaces are to
each other as the squares of their like dimensions,
we have

Fig. 538

trl. FBA : trl.

I'

' I' /

Ul,

Fig. 539

DHB : trl. EAH = AB® ; BH®

: AH^, whence trl. PBA : trl.
DHB + trl. EAH = AB® : BH®

+ AH®. But trl. FBA = trl.
DAH + trl. EAH. .*. AB® = BH®

+ AH®.

.*. sq. upon AB = sq.
upon HD + sq. upon HA.

a. Devised by the
author Sept. l8, 1900, for
similar regular polygons
other than squares.

In fig. 559 > from the
middle points of AB, BH and
HA draw the three perp*s PE,
GC and KD, making PE = 2AB,

GEOMETRIC PROOFS

237

GC = 2BH and KD = 2HA, complete the three isosceles
trl‘s EBA, CHB and DAH, and draw EH, BK and DB.

Since these trl*s are respectively equal to
the three sq's upon AB, BH and HA, it remains to
prove tri. EBA = trl. CHB + tri. DAH. The proof is
same as that in fig. 338, hence proof for 339 is a
variation of proof for 338.

a. Devised by the author, because of the fig-
ure, so as to get area of trl. EBA = AB^, etc. AB^
= BH^ + AH^.

.*. sq. upon AB = sq, upon BH + sq. upon AH.

/. h^ = a^ + b^.

b. This proof is given by Joh. Hoffmann; see
his solution in Wipper*s Pythagoralsche Lehrsatz,
l880, pp. -4*5-48.

See, also, Beman and Smith* s New Plane and
Solid Geometry, 1899, p. 105, ex. 207; Versluys, p.

59, fig. 63.

c. Since any polygon of three, four, five,
or more sides, regular or irregular, can be trans-
formed, (see Beman and Smith, p, 109), into an equiv-
alent triangle, and it into an equivalent isosceles
triangle whose base is the assumed base of the poly-
gon, then is the sum of the areas of two such similar
polygons, or semicircles, etc., constructed upon the
two legs of any right triangle equal to the area of

a similar polygon constructed upon the hypotenuse of
said right triangle, ^ the sum of the two Isosceles
triangles so constructed, (be their altitudes what
they may), is equal to the area of the similar isosce-
les triangle constructed upon the hypotenuse of the
assumed triangle. Also see Dr. Leltzmann, ( 1930 ),
p. 37, fig. 36 for semicircles.

d. See proof Two Hundred Forty-One for the
establishment of above hypothesis.

Iwflt-HiLalLSLl.EftLtl-fine

Let trl *3 CBA, DHB and EAH be similar isos-
celes trl*s upon the bases AB, BH and AH of the rt.

238

THE POTHAGOREAN PROPOSITION

Fig. 3*^0

paral. E'G'QB
paral. HBRT =
same as paral.

trl. ABH, and CP, DG and EK their
altitudes from their vertices 0,

D and E, and L, M and N the mid-
dle points of these altitudes.

Transform the trl's DHB,
EAH and CBA Into their respective
paral 's BRTH, AHUW and OQBA.

Produce RT and WO to X,
and draw XHY. Through A and B
draw A 'AC and ZBB' par. to XY.
Through H draw HD' par. to OQ
and complete the paral. HP'. Draw
XD ' and E ' Z . Trl ' s E ' YZ and XHD
are congruent, since YZ = HD' and
respective angles are equal. EY
= XH. Draw E'G' par. to BQ, and
= paral. E'YZB = paral. XHD'P; also
paral. HBB'X. But paral. HBB'X Is
XHBB' which = paral. XHD'P' = paral.

E'YZB.

.'. paral. E'G'QB = trl. DHB; In like manner
paral. AOG'E' = trl. EAH. As paral. AA'ZB = paral.
AOQB = trl. CBA, so trl. CBA = trl. DHB + trl. EAH.
(l).

Since trl. CBA : trl. DHB : trl. EAH = h® : a®
: b®, trl. CBA : trl. DHB + trl. EAH = h® : a^ + b®.

But trl. CAB = trl. DHB + trl. EAH. (l). h®

= a^ + b®.

sq. upon AB = sq. upon BH 4- sq. upon AH.

Q.E.D.

a. Original with author. Formulated Oct. 28,
1933 . The author has never seen, nor read about, nor
heard of, a proof for h^ = a^ + b^ based on Isosceles
triangles having any altitude or whose equal sides
are unrelated to a, b, and h.

Let X, Y and X be three similar pentagons on
sides h, a and b. Then, if X = Y + Z, h^ = a^ + b^.

GEOMETRIC PROOFS

239

Transform pentagon X, Y and Z into equivalent
trl’s DQO, RGT and UMVT. Then, (by 4th proportional.
Pig. a), transform said trl*s Into equivalent Isosce-
les trl»s P*BA, S*HB and V*AH.

Then proceed as In fig. 340. .-. h^ = a® + b^.

sq. upon AB = sq. upon BH + sq. upon AH. Q.E.D.

Or using the similar trl*s XBH, YHB and ZAH,
proving tri. XAB = trl. YHB + trl. ZAH, whence 5 trl.
XBA = 5 trl. YHB + 5 trl. ZAH; etc.

240

THE PYTHAGK)REAN PROPOSITION

By argiunent established under flg*3 540 and
34l, if regular polygons of any number of sides are
const *d on the three sides of any rt. triangle, the
sum of the tvo lesser = the greater, whence always
h^ = a^ + b^.

a. Devised by the author, Oct. 29, 1935*

b. In fig. a, 1-2 = HB; 2-5 = TR; 1-4 = OS;
4-5 = SS'; 1-B = AH; 6-7 = WU; 1-8 = MV; 8-9 = VV* ;
1-10 = AB; 11-11 = OQ; 1-12 = PD; 12-15 = P'P.

trl.

In fig. 542, produce AH to
E making HE = HB, produce BH to P
making HP = HA, draw RB perp. to AB
making BK = BA, KD. par. to AH, and
draw EB, KH, KA, AD and AP. BD = AB
and KD = HB.

Area of trl. ABK = (area of
trl. KHB = area of trl. EHB) + (area
of trl. AHK = area of trl. AHD)

+ (area of ABH = area of ADP).
area of ABK = area of trl, EHB + area of
AHP. sq. upon AB = sq. upon BH + sq. upon AH,
= a^ + b^.

a. See Edwards* Geom. , 1895# P*

158, fig. (20).

Fig. 3^5

+ Jed X db

In fig. 343, take AD = AH,
draw ED perp. to AB, and draw AE.

Trl. ABH and BED are similar, whence
DE = AH X BD -e- HB. But DB = AB - AH.
Area of trl. ABH = JAH x bH
AD X ED
T.

= 2

+ iED X dB

_ AH^CAB - AH) i AH (MB - AH)
BH ® BH

AD

2

ED

BH"^

= 2AH X AB - 2AH^ + AB^ + AH^
= BH^ + AH^.

2AH X AB.

AB**

GfEOMETRIC PROOFS

241

.% sq. upon AB = sq. upon BH + sq, upon AH.
h® = a® + b®.

a. See Am. Math. Mo., Vol. VI, l899> P- 70,
proof XCV.

h. See proof Five , fig. 5s under I, Algebraic
Proofs, for an algebraic proof.

In fig. 544, produce
BA to L making AL = AH, at L
draw EL perp. to AB, and pro-
duce BH to E. The trl’s ABH
and EBL are similar.

Area of trl. ABH = JAH
X BH = JLE X lb - LE X la

1 AH(AH + AB)^ _ AH^ (AH + AB )
^ BH BH '

whence AB® = BH® + AH®.

.'. sq. upon AB = sq.
upon BH 4- sq. upon AH. .% h®

= a + b .

a. See Am. Math. Mo.,
Vol. VI, 1899, P. 70, proof
XCVI.

b. This and the preced-
ing proof are the converse of each other. The two
proofs teach that if two triangles are similar and so
related that the area of either triangle may be ex-
pressed principally In terms of the sides of the
other, then either triangle may be taken as the prin-
cipal triangle, giving, of course, as many solutions
as It Is possible to express the area of either In
terms of the sides of the other.

! \

* \

^ N
* \

• \

* \

* \

I \

I \

\

\

\

• Fig. 544

In fig. 545, produce HA and HB and describe
the arc of a circle tang, to HX, AB and HY. Prom 0,

242

the PYTHAGOREAN PROPOSITION

Pig. 345

j. -kS

= a + D .

/. sq. upon AB = sq.

its center, draw to
points of tangency, OG,
OE and OD, and draw OH.

Area of sq. DG
o 1 / BE ^ r

= r^ = iab + (2 r

^ AE X r^ 1 , . ,

+ 2 2 / “

But since 2 r=h+a+b,
r = ^ (h + a + b ) .

-J- (h + a + b )^ = Jab
+ h(h + a + b), whence

upon BH + sq. upon AH.

a. This proof is original with Prof. B. F.
Yanney, Wooster University, 0. See Am. Math. Mo.,

voi. VI, 1899, p. 70, XCVII.

Seven

In fig. 5^6, let AE
= BH. Since the area of a
circle is Tcr^, if it can be
proven that the circle whose
radius is AB = the circle
whose radius AH + the circle
whose radius is AE, the truth
sought is established.

It is evident, if the
triangle ABH revolves in the
plane of the paper about A as
a center, that the area of
the circle generated by AB
will equal the area of the
circle generated by AH plus the area of the annulus
generated by HP.

Hence it must be shown, if possible, that the
area of the annulus is equal to the area of the cir-
cle whose radius is AE.

GfEOMETRIC PROOFS

24 ?

Let AB = h = AF, AH = b, BH = a, AD = ^BH = r, HK

= KF, and AK = mr, whence GH = h + h, AK = — = mr,

HF = h-b, HK=KF = - — .

Now (GH = h + b) : (BH = 2r) = (BH = 2i*)

: (HF = h - b). (l)

whence h = v/b® + and b = \/h® - 4r®, /. ^

\/b® + + b / I. h + b

= * r = mr, whence b = r (m - — and — r —

2 m 2

h + \/h 2 - 4r2 , , / . 1 , a - b

= * r = mr, whence h = r(m + /. — r —

r(m + i) - r(m - i)

= “ = HK.
m

Now since (AD = r)

: (AK = mr) = (HK = : (AD = v). —(2)

A AD ; AK = HF : AE, or 2nAD : 2nAK = HF : AE,

A 2nAK X hF = 2nAD x aE, or 2it ( - | - - )HF = nAE x AE.

But the area of the annulus equals J the sum
of the clrciomferences where radii are h and b times
the width of the annulus or HF.

/. the area of the annulus HF = the area of
the circle where radius is HB,

the area of the circle with radius AB = the
area of the circle with radius AH + area of the an-
nulus ,

nh^ = Tia^ 4- Tib^.

sq. upon AB = sq. upon BH + sq. upon AH.

A h® = a® + b®.

a. See Am. Math. Mo., Vol. I, 1894, p. 22?,
the proof by Andrew Ingraham, President of the Swain
Free School, New Bedford, Mass,

b. This proof, like that of proof Two Hundred
Fifteen , fig. 513 proves too much, since both equa-
tions (l) and (2) imply the truth sought. The author.
Professor Ingraham, does not show his readers how he

r

determined that HK = — , hence the implication is

m

hidden; in (l) we have directly h^ - b® = (4r^ = a^).

244

THE PYTHAGOREAN PROPOSITION

Having begged the question In both equations,
(l) and (2), Professor Ingraham has, no doubt, un-
consciously, fallen under the formal fallacy of
petttio principi i .

c. From the preceding array of proofs It Is
evident that the algebraic and geometric proofs of
this most Important truth are as unlimited In number
as are the Ingenious resources and Ideas of the mathe
matlcal Investigator.

NO TRIGONOMETRIC PROOFS

Facing forward the thoughtful reader may
raise the question: Are there any proofs based upon
the science of trigonometry or analytical geometry?

There are no trigonometric proofs, because
all the fundamental formulae of trigonometry are them
selves based upon the truth of the Pythagorean Theo-
rem; because of this theorem we say sln^A + cos^A
= 1, etc. Trlglnometry is because the Pythagorean
Theorem t s.

Therefore the so-styled Trigonometric Proof,
given by J. Versluys, In his Book, Zes en Negentlg
Bewljzen, 1914 (a collection of 96 proofs), p, 94,
proof 95s Is not a proof since It employs the formula
sln^A + cos^A = 1.

As Descartes made the Pythagorean theorem the
basis of his method of analytical geometry, no Inde-
pendent proof can here appear. Analytical Geometry
Is Euclidian Geometry treated algebraically and hence
Involves all principles already established.

Therefore In analytical geometry all rela-
tions concerning the sides of a right-angled triangle
Imply or rest directly upon the Pythagorean theorem
as Is shown In the equation, viz,, + y^ = r^.

And The Calculus being but an algebraic In-
vestigation of geometric variables by the method of
limits It accepts the truth of geometry as estab-
lished, and therefore furnishes no new proof, other
than that. If squares be constructed upon the three

RENE DESCARTES
1596-1650

Hi;

GEOMETRIC PROOFS

245

sides of a variable oblique triangle, as any angle of
the three approaches a right angle the square on the
side opposite approaches in area the sum of the
squares upon the other tvo sides.

But not so with quaternions, or vector analy-
sis. It is a mathematical science which Introduces
a new concept not employed in any of the mathematical
sciences mentioned heretofore, — the concept of direc-
tion.

And by means of this new concept the complex
demonstrations of old truths are wonderfully simpli-
fied, or new ways of reaching the same truth are de-
veloped.

III. QUATERNIONIC PROOFS

We here give four quaternionio proofs of the
Pythagorean Proposition. Other proofs are possible.

Sine

In fig. 547 designate the
sides as to distance and direction by
a, b and g (in place of the Greek al-
pha a, beta p and gamma y). Now, by
the principle of direction, a = b + g;
Fig. 347 also since the angle at H is a right
angle, 2sbg = 0 (s signifies Scalar.
See Hardy, I88I, p. 6).

(1) a + b = g (if = (2) a^ = b® + 2sbg + g^
(2) reduced = (3). a® = b® + g^, considered as
lengths. /. sq. upon AB = sq. upon BH + sq. upon AH.
h^ = a^ + b^. Q.E.D.

a. See Hardy’s Elements of Quaternions, l88l,
p. 82, art. 54, 1; also Jour, of Education, Vol.
XXVII, 1888, p. 327, Twenty-Second Proof; Versluys,

p. 95, fig. 108.

Iwa

In fig. 548, extend BH to C
making HC = HB and draw AC. As vec-
tors AB=AH+HB, orA=B+G (l).
Also AC = AH + HC, or A = B - G (2).

Squaring (l) and (2) and
adding, we have A® + A^ = 2B® + 2G®.
Or as lengths, AB® + AC® = 2AH®

Fig. 548 + 2AB®. But AB = AC.

AB® = AH® + HB®.

sq. upon AB = sq. upon AH + sq. upon HB.

.'. h® = a® + b®.

246

QUATERNIONIC PROOFS

247

a. This Is James A. Calderhead* s solution.
See Am. Math. Mo., Vol. VI, 1899, p. 71, proof XCIX.

Three

Fig. 349

In fig. 349 , complete the rect.
HC and draw HC. As vectors AB = AH
+ HB, or a = b + g (l ). HC = HA + AC,
or a = -b + g ( 2 ).

Squaring (l ) and (2 ) and add-
ing, gives A^ + A*^ = 2B^ + 2G^. Or
considered as lines, AB^ + HC^ = 2AH^

+ 2HB^. But HC = AB.

AB^ = AH^ + HB^.

sq. upon AB = sq. upon AH^ + sq. upon HB^.

a. Another of James A. Calderhead's solutions.
See Am. Math. Mo., Vol. VI, 1899, p. 71, proof C;
Versluys, p. 95, fig. IO 8 .

\ f' ^ y /

Fig. 350

In fig. 550 , the con-
struction is evident, as an-
gle GAK = -angle BAR. The ra-
dius being unity, LG and LB
are sines of GAK and BAK.

As vectors, AB = AH
+ HB, or a = b + g (l ) . Also
AG = AP + PG or a' =— b + g
( 2 ). Squaring (l ) and (2)
and adding gives a^ + a*^

= 2b^ + 2g^. Or considering
the vectors as distances, AB^
+ AG^ = 2AH^ + 2HB^, or AB^

AH^ + HB^.

sq. upon AB = sq. upon AH + sq. upon BH.
h® = a^ + b^.

a. Original with the author, August, 1900.

b. Other solutions from the trigonometric
right line function figure (see Schuyler *s Trigonome-
try, 1873 , p. 78 , art 85 ) are easily devised through
vector analysis.

IV. DYNAMIC PROOFS

The Science of Dynamics, since 1910, is a
claimant for a place as to a few proofs of the Pytha-
gorean Theorem.

A dynamic proof employing the principle of
moment of a couple appears as proof 96, on p. 95 > In
J, Versluys* (191^) collection of proofs.

It Is as follows:

do.®

In compliance with the
theory of the moment of couple.
In mechanise (see ^Mechanics
for Beginners, Part I," I89I,
by Rev. J. B. Locke, p. 105),
the moment of the sum of two
conjoined couples In the same
flat plane Is the same as the
sum of the moments of the two
couples, from which it follows
that h^ = a^ + b^.

If PH and AG represent
two equal powers they form a
couple whereof the moment
equals PH AH, or b^.

If HE and DB represent two other equal powers
they form a couple whereof the moment equals DB x HB
or a^.

To find the moment of the two couples join
the two powers AG and HE, also the two powers DB and
PH. To join the powers AG and HE, take AM = HE. The
diagonal AN of the parallelogram of the two powers AG
and AM Is equal to CA. To join the powers PH and DB,
take BO = DB. The diagonal BK of the parallelogram
of the two powers (PH = BP ) and BO, Is the second

DYNAMIC PROOFS

2^9

component of the resultant couple whose moment is CH
X BK, or h^. Thus we have h^ = a^ +

a. See J. Versluys, p. 95 , fig. 108. He
(Versluys ) says: I found the above proof In 1877^
considering the method of the theory of the principle
of mechanics and to the present (l91^) I have never
met with a like proof anywhere.

In Science, New Series, Oct. 7 , 1910, Vol.52,
pp. 865-4, Professor Edwin F. Northrup, Palmer Physi-
cal Laboratory, Princeton, N.J., through equilibrium
of forces, establishes the formula h^ = a^ + b^.

In Vol. 33, p. 457, Mr. Mayo D. Hersey, of
the U.S, Bureau of Standards, Washington, D.C., says
that, if we admit Professor Northrup* s proof, then
the same result may be established by a much simpler
course of reasoning based on certain simple dynamic
laws .

Then in Vol. 34, pp. l8l-2, Mr. Alexander Mac-
Far lane, of Chatham, Ontario, Canada, comes to the
support of Professor Northrup, and then gives two
very fine dynamic proofs through the use of trigono-
metric functions and quaternionlc laws.

Having obtained permission from the editor of
Science, Mr. J. McK. Cattell, on February l8, 1926,
to make use of these proofs found in said volumes 32,
33 and 34, of Science, they now follow.

In fig. 352, 0-p is a rod
without mass which can be revolved
in the plane of the paper about 0
as a center. 1-2 is another such
rod in the plane of the paper of
which p is its middle point. Con-
centrated at each end of the rod
1-2 are equal masses m and m*
each distant r from p.

Let R equal the distance
0-p, X = 0-1, y = 0-2. When the

Fig. 352

250

THE PYTHAGOREAN PROPOSITION

system revolves about 0 as a center, the point p will
have a linear velocity, r = ds/dt = da/dt = R¥, where
ds Is the element of the arc described In time dt, da
Is the differential angle through which 0-p turns,
and W Is the angular velocity.

1. Assume the rod 1-2 free to turn on p as a

center. Since m at 1 and m* at 2 are equal and equal-
ly distant from p, p Is the center of mass. Under
these conditions E* = |•(2m)V^ = mR^W^. (l )

2. Conceive rod, 1-2, to become rigorously
attached at p. Then as 0-p revolves about 0 with an-
gular velocity W, 1-2 also revolves about p with like
angular velocity. By making attachment at p rigid
the system Is forced to take on an additional kinetic
energy, which can be only that, which Is a result of
the additional motion now possessed by m at 1 and by
m* at 2, In virtue of their rotation about p as a
center. This added kinetic energy Is E” = |■(2m)r^W^

= mr^W^. (2) Hence total kinetic energy Is E = E*

+ E" = mW^(R^ + r^). — -(?)

3. With the attachment still rigid at p, the
kinetic energy of m at 1 Is, plainly, E© = ■g-mx^W^.

(4) Likewise Eq = gmy^W^. (5)

/. the total kinetic energy must be E = E©

+ Eq = ^mW^(x^ + y^). (6)

(3) = (6), or i(x^ + y^) = R^ + r^ —(7)

In (7) we have a geometric relation of some
Interest, but In a particular case when x = y, that
is, when line 1-2 Is perpendicular to line 0-p, we
have as a result x^ = R^ + r^. (8)

/. sq. upon hypotenuse = sum of squares upon
the two legs of a right triangle.

Then In Vol. 33, P- ^57, on March 24, 1911,

Mr. Mayo D. Hersey says: "while Mr. R. P. Delmal
holds that equation (7) above expresses a geometric
fact — I am tempted to say ’accident* — which textbooks
raise to the dignity of a theorem." He further says:
"Why not let It be a simple one? For Instance, If
the force P whose rectangular components are x and y,
acts upon a particle of mass m until that v^ must be

DYNAMIC PROOFS

251

positive; consequently, to hold that the square of a
simple vector is negative is to contradict the estab-
lished conventions of mathematical analysis.

The quaternlonist tries to get out by saying
that after all v is not a velocity having direction,
but merely a speed. To this I reply that E = cos
y*mvdv = 2 niv^, and that these expressions v and dv are
both vectors having directions which are different.

Recently (in the Bulletin of the Quaternion
Association) I have been considering what may be

called the generalization of the Pytha-
gorean Theorem.

Let A, B, C, D, etc., fig. 355,
denote vectors having any direction in
space, and let R denote the vector from
the origin of A to the terminal of the
last vector; then the generalization of
the P.T. Is R^ = A^ + + C^ +

+ 2 (cos AB + cos AC + cos AD) + 2 (cos BC
353 + cos BD) + 2 (cos CD) + etc., where cos

AB denotes the rectangle formed by A and
the projection of B parallel to A. The theorem of P.
is limited to two vectors A and B which are at right
angles to one another, giving = A^ + B^. The ex-
tension given in Euclid removes the condition of per-
pendicularity, giving R^ = + B^ + cos AB,

Space geometry gives R^ = A^ 4- B^ + when
A, B, C are othogonal, and R^ = A^ + B® + C^ + 2 cos
AB + 2 cos AC + 2 cos BC when that condition is re-
moved.

Further, space-algebra gives a complementary
theorem, never dreamed of by either Pythagoras or
Euclid.

Let V denote in magnitude and direction the
resultant of the directed areas enclosed between the
broken lines A + B + C + D and the resultant line R,
and let sin AB denote in direction and magnitude the
area enclosed between A and the projection of B which
is perpendicular to A; then the complementary theorem
la 4V = 2 (sin AB + sin AC + sin AD + ) + 2 (sin BC

+ sin BD + ) + 2 (sin CD + ) + etc.

252

THE PYTHAGOREAN PROPOSITION

THE PYTHAGOREAN CURIOSITY

The following Is reported to have been taken
from a notebook of Mr. John Waterhouse, an engineer

of N.y. City. It
appeared In print.
In a N.Y. paper.

In July, 1899.

Upon the sides of
the right tri-
angle, fig. 354,
construct the
squares AI, BN,
and CE. Connect
the points E and
H, I and M, and
N and D. Upon
these lines con-
struct the squares
EG, MK and NP,
and connect the
points P and P, G
and K, and L and

0. The following
truths are demon-
strable .

1 . Square
BN = square CE
+ square AI. (Eu-
clid).

2. Triangle HAE = triangle IBM = triangle DCN
= triangle CAB, since HA = BI and EA = MY, EA = DC
and HA = NZ, and HA = BA and EA = CA.

3. Lines HI and GK are parallel, for, since
angle GHI = angle IBM, /. triangle HGI = triangle BMI,
whence IG = IM = IK. Again extend HI to H* making

IH* = IH, and draw H*K, whence triangle IHG = triangle
IH*K, each having two sides and the Included angle
respectively equal. /. the distances from G and K to
the line HH* are equal. /. the lines HI and GK are
parallel. In like manner It may be shown that DE and
PP, also MN and LO, are parallel.

DYNAMIC PROOFS

253

4. GK = 4HI, for HI = TU = GT = UV = VK
(since VK Is homologous to BI in the equal triangles
VKI and BIM). In like manner it can he shown that
PP = 4DE, That LO = 4MN is proven as follows: tri-
angles LWM and IVK are equal; therefore the homolo-
gous sides WM and VK are equal. Likewise OX and QD
are equal each being equal to MN. Now in trl. WJX,

MJ and XN = NJ; therefore M and N are the middle
points of WJ and XJ; therefore WX = 2MN; therefore
LO = 4MN.

5. The three trapezoids HIGK, DEPP and, MNLO
are each equal to 5 times the triangle CAB. The 5
triangles composing the trapezoid HIGK are each equal
to the triangle CAB, each having the same base and
altitude as triangle CAB. In like manner it may be
shown that the trapezoid DEPP, so also the trapezoid
MNLO, equals 5 times the triangle CAB.

6. The square MK + the square NP = 5 times
the square EG or BN. For the square on MI = the
square on MY + the square on YI + (2AB)^ + AC^ = 4AB^
+ AC^; and the square on ND + the square on NZ + the
square ZD = AB^ + (2AC)^ = AB^ + 4AC^. Therefore the
square MK + the square NP = 5AB^ + 5AC^ = 5(AB^+AC^)

= 5BC^ = 5 times the square BN.

7. The bisector of the angle A’ passes through
the vertex A; for A*S = A*T. But the bisector of the
angle B* or C, does not pass through the vertex B,

or C. Otherwise BU would equal BU', whence NU'* + U"M
would equal NM + U"M* ; that is, the sum of the two
legs of a right triangle would equal the hypotenuse
+ the perpendicular upon the hypotenuse from the right
angle. But this is Impossible. Therefore the bisec-
tor of the angle B* does not pass through the vertex
B.

8. The square on LO = the s\am of the squares
on PP and GK; for LO : PP : GK = BC : CA : AB.

9. Etc., etc.

See Casey’s Sequel to Euclid, 1900, Part I,

p. 16 .

PYTHAGOREAN MAGIC SQUARES

fio.®

The Siam of any row,
column or diagonal of the
square AK is 125; hence the
sum of all the numbers in the
square is 625. The sum of any
row, column or diagonal of
square GH is 46, and of HD is
147; hence the sum of all the
numbers in the square GH is
l84, and in the square HD is
441. Therefore the magic
square AK (625) = the magic
square HD (44l ) + the magic
square HG (l84 ) .

Formulated by the author, July, I9OO.

Iwg

fig. 556

The square AK is com-
posed of 5 magic squares, 5^,
15 ^ and 25 ^. The square HD is
a magic square each number of
which is a square. The square
HG is a magic square formed
from the first I 6 niombers. Fur-
thermore, observe that the sum
of the nine square numbers in
the square HD equals 48^ or
2504 , a square number.

Formulated by the
author, July, 19OO.

254

PYTHAGOREAN MAGIC SQUARES

255

Ihree

The Siam of all the num-
bers (AK = 525) = the sum of
all the numbers in square (HD
= 189) + the sum of all the
numbers In square (HG + I36).

Square AK Is made up
of 13, 3 X (3 X 13), and 5
(5 X 15); square HD is made
up of 21, 3 X (3 X 21), and
square HG is made up of ^ x 34
- each row, column and diag-
onal, and the sum of the four
Fig. 357 inner numbers.

Many other magic squares

of this type giving 325, I89 and I36 for the sums of
AK, HD and HG respectively may be formed.

This one was formed by Prof. Paul A. Towne,
of West Edmeston, N.Y.

EftyiL

type may be formed,
my own of this type.

The sum of numbers in
sq. (AK = 625) = the sum of
numbers in sq. (HD = 44l) -f the
sum of numbers in sq. (HG
= 184).

Sq. AK gives 1 x (1

X 25); 5 X (3 X 25); and 5

(5 X 25), as elements; sq. HD
gives IX (1 X 49); 3 X (3 X 49)
as elements; and sq. HG gives
1 X 46 and 3 ^ ^6, as elements.

This one also was formed
by Professor Towne, of West
Edmeston, N.Y. Many of this
See fig. 355, above, for one of

256

THE PYTHAGOREAN PROPOSITION

Also see Mathematical Essays and Recreations,
by Herman Schubert, In The Open Court Publishing Co,,
Chicago, 1898 , p. 39 ^ Por an extended theory of The
Magic Square .

Five

Observe the following series:

The Siam of the Inner 4 numbers is 1^ x 202;
of the 16 -square, 2^ X 202; of the 36 -square, 3^

X 202; of the 64- square, 4^ x 202; and of the 100-
square, 5^ X 202.

SSXEDOIEEBSSIDDZ

II 31 S3l[Z B R3 3! !! 9
'nsfriSiprmsisEffr
acRSRirBSsase

n E7i ’n ^ nt SI SK rr 7>

assssiRifigaailSB

'SlDCSIKtSJBiliiUQlZ

Fig. 559

"On the hypotenuse and legs of the right-
angled triangle, ESL, are constructed the concentric
magic squares of 100, 64, 36 and I 6 . The sum of the
two numbers at the extremities of the diagonals, and

PYTHAGOREAN MAGIC SQUARES

257

of all lines, horizontal and diagonal, and of the two
numbers equally distant from the extremities. Is 101,
The sum of the numbers In the diagonals and lines of
each of the four concentric magic squares Is 101 mul-
tiplied by half the number of cells In boundary lines;
that Is, the summations are 101 x 2; 101 x lOl x 4;
101 X 5. The sum of the 4 central numbers Is 101 x 2,

/. the sum of the numbers In the square (SO
= 505 ^ 10 = 5050) = the sum of the numbers In the
square (EM = 303 x 6 = I818) + the sum of the numbers
In the square (El = 404 x 8 = 3232). 505® = 303®

+ 404 ®.

Notice that In the above diagram the concen-
tric magic squares on the legs Is Identical with the
central concentric magic squares on the hypotenuse."
Professor Paul A. Towne, West Edmeston, N.Y.

An Indefinite number of magic squares of this
type are readily formed.

ADDENDA

The following proofs have come to
me since June 25, 1939, the day on which I
finished page 257 of this 2nd edition.

= sq. BE + 2 trl. BHG)
agon GABDEP - 2 trl. ABH = sq. AP +
fore sq. upon AB = sq. upon HB + sq

In fig. 360 , extend
HA to P making AP = HB, and
through P draw PQ par. to HB,
making CQ = HB; extend GA to
0, making AO = AG; draw PE,

GE, GD, GB, CO, QK, HC and BQ.

Since, obvious, trl.
KCQ = tri. ABC = trl. PEH,
and since area of tri. BDG
= ^BD X PB, then area of quad.
GBDE = BD X (PB = HP) = area
of paral. BHCQ = sq. BE + 2
trl. BHG, then it follows
that :

Sq. AK = hexagon
ACQKBH - 2 trl. ABH = (trl.

ACH = trl. GAB) + (paral. BHCQ
+ (trl. QKB = trl. GPE) = hex-

sq. BE.
upon HA.

There-

= a® + b®, Q.E.D.

a. Devised, demonstrated with geometric rea-
son for each step, and submitted to me June 29, 1939.
Approved and here recorded July 2, 1939, after ms.
for 2nd edition was con 5 )leted.

b. Its place, as to type and figure. Is next
after Proof Slxty-Nlne . p. l4l, of this edition.

c. This proof Is an Original, his No. VII, by
Joseph Zelson, of West Phlla. High School, Phlla., Pa.

258

ADDENDA

259

Ittft-!!.u.Q.iLS.4_E. ftLlY.-fi.Ias.

Fig. 561

aq. upon AB =
h® = a* + b^.

It is easily proven
that: trl. OPY = trl. PHE
= trl. ABH = trl, CMA = trl.
ARC = trl. CWX; also that
trl. QAE = trl. CMH; trl.

LGE = trl. CXHj trl. PYL
= trl, EDN = trl, WKY; trl.
GPY = trl. GPL + trl. NEDj
that paral. BHVfK = aq. HD.
Then It follows that sq. AK
= pentagon MCKBH - 2 trl,

ABH = (trl. MCH = trl. GAE)

+ (trl. CXH = trl, LGE)

+ [(quad. BHXK = pent. HBDHE)
= sq. BE + (trl. EDN = trl.
WKX)] = hexagon GAHBDNL - 2
trl. ABH = sq. AP + sq. BE.
sq. upon HB + sq. upon HA.

a. This proof, with figure, devised by Mas-
ter Joseph Zelson and submitted June 29, 1939, and
here recorded July 2, 1939,

b. Its place Is next after No. 247, on p.
185 above.

Iaft_tlu.alLft4_ELttY

Fig. 362

In fig. 362, draw GD.
At A and B erect perp's AC
and BK to AB. Through L and
0 draw PM, and EN = PM = AB.
Extend DE to K.

It Is obvious that:
quad. GMLC = quad. OBDEj
quad. OBDE + (trl. LMA = trl.
OKE) = trl. ABH; trl, BDK
= trl. EDN = trl. ABH = trl,
EPH = trl. MPG = trl. CAG.

260

THE PYTHAGOREAN PROPOSITION

Then it follows that: sq. GH + sq. HD = hex-
agon GABDEP - 2 trl. ABH

trap . FLOE

(PL + OE = FM = AB) X (FE = AB ]
2

+ ^trap. LABO ^ ~ ^ = AB^

= sq. on AB.

/. sq. upon AB = sq. upon HB + sq. upon HA.

. . h = a + D .

a. Type J, Case (l), (a). So Its place Is
next after proof Two Hundred Nine, p. 2l8.

b. Proof and fig. devised by Joseph Zelson.
Sent to me July 13, 1959-

Iwa^Hiindred^F j.f ty-Jnfe

Construct trl. KGP
= trl. ABH; extend PE to L
\ r* point at which a

j perp. from D Intersects PE

! \ n ^x'tended; also extend AB to

Gk^ ^ ^ ^ \ 7 M N where perp * s from G

^ will Intersect AB ex-

[ \ tended; draw GD,

IM m! By showing that:

tri. KLF = tri. DOE = trl.
Fig. 363 DNBj trl. PLG = trl. AMG;

trl. KGP = trl. EPH = trl.

ABH; then It follows that: sq. (31 + sq. HD = hexagon
LGMNDO - 4 trl. ABH

Fig. 363

trap . LGDO =

|trap .

- 4 trl. ABH =

(LG+D0 = KG = AB) x (PE = AB)+(2 x alt.PL)
2

(GM+HD= AB)x(ab)+( 2 X alt. {aM = PL> )1

- 4 trl. ABH = AB® - 4 trl. ABH.

sq. upon AB = sq. upon HB + sq. upon HA.
h® = a® + b®. (a.E.D.

ADDENDA

261

a. Type J. Case (l), (a). So Its place is
next after Proof Two Hundred Fifty-One,

b. This proof and fig. also devised by Master
Joseph Zelson, a lad with a superior Intellect. Sent
to me July 15, 1939.

By dissection, as per
figure, and the numbering of
corresponding parts by same
numeral, it follows, through
X J ^ 'x o V superposition of congruent

parts (the most obvious proof )
I ^ that the sum of the four

\ 1 parts (2 trl's and 2 quad' Is)

in the sq. AK = the sum of
the three parts (2 trl*s and

Fig. 364 1 quad. ) in the sq. PC + the

sum of the two parts (l trl.
and 1 quad. ) in the sq. PD.

That is the area of the sum of the parts 1+2
+ 5 + ^ in sq. AK (on the hypotenuse AB ) = the area
of the sum of the parts 1’ + 2* + 6 in the sq. PC
(on the line GP = line AH) + the area of the sum of
the parts 5* +4* in the sq, PD (on the line PK
= line HB), observing that part 4 + (6 = 5) = part 4*.

a. Type I, Case (6), (a). So its place be-

longs next after fig. 305 , page 215.

b. This figure and proof was devised by the
author on March 9, 19^0, 7^30 p.m.

Fig. 564

and 1 quad. ) in the sq.

Iw&.tLu,ndred_Fift^- Three

In "Mathematics for the Million," (1937), 1)7
Lancelot Hogben, P.R.S., from p. 65, was taken the
following photostat. The exhibit is a proof which is
credited to an early (before 500 B.C.) Chinese mathe-
matician. See also David Eugene Smith* s History of
Mathematics, Vol. I, p. 50.

262

THE PYTHAGOREAN PROPOSITION

Mathematics in Prehistory

Fig. 19

The Book of Chou Pei Suan King, probably written about A D. 40, is
attributed by oral tradition to a source before the Greek geometer taught what
we call the Theorem of Pythagoras, i.e. that the square on the longest side of
a right-angled triangle is equivalent to the sum of the squares on the other
t^^o. This very early example of block printing from an ancient edition of the
Chou Peti as gi\ cn in Smith’s History of MathematicSy demonstrates the truth
of the theorem. By joining to any right-angled triangle like the black figure
«Bf three other right-angled triangles just like it, a square can be formed.
Next trace four oblongs (rectangles) like eu/B, each of which is made up of
two triangles like e/B. When you have read Chapter 4 you will be able to pul
together the Chinese puzzle, which is much less puzzling than Euclid. These
are the steps :

Triangle «/B = } rectangle ea/B = | B/ .

Square ABCD = Square efgh -h 4 times triangle e/B
- eft -f 2B/.eB

Also Square ABCD B/* -f eB® + 2B/ . eB
So «/* -f- 2B/.eB . B/* } eB* + 2B/.eB

Hence «/• = B/* + eB-

a. This belleve-lt-or-not "Chinese Proof" be-
longs after proof Ninety , p, 154, this book. (E.S.L.,
April 9, 1940).

GEOMETRIC PROOFS

263

= sq
=

, HD) +
+ b*

In the figure extend
GP and DE to M, and AB and
ED to L, and number the
parts as appears in the
quad. ALMG.

It la easily shown
that: AABH = AACG, ABKN
= AKBL and ACNP = AKOEj
whence DaK = (AABH = AACG
In sq. HG) + quad. AHNC com.
to D'a AK and HG + (ABKN
= AKBL) = (ABLD + quad. BDEO
(AOEK = ANPC) = DHD + DHG. Q.E.D.

a. This fig. and demonstration was formulated
by Fred. W. Martin, a pupil In the Central Junior-
Senior High School at South Bend, Indiana, May 27,
19 ^ 0 .

b. It should appear in this book at the end
of the B-Type section. Proof Ninety-Two.

Draw CL perp. to AH, join
CH and CE; also G®. Construct
sq. HD' = sq. HD. Then observe
that ACAH = ABAC, ACHE = AABH,
ACER = ABPG and AMEK = ANE'A.

Then It follows that sq.
AK = (AAHC = AAG© In sq. HG)

+ (AHEC = ABHA In sq. AK) + (i^EKC
= ABPG In sq. HG) + (ABDK - AMEK
= quad. BDEM In sq. HD) + AHBM
com. to sq's AK and HD = sq. HD
+ sq. HG. /. h® = a® + b*.

a. This proof was dis-
covered by Bob Chlllag, a pupil

264

THE PYTHAGOREAN PROPOSITION

In the Central Junior-Senior High School, of South
Bend, Indiana, in his teacher’s (Wilson Thornton’s)
Geometry class, being the fourth proof I have re-
ceived from pupils of that school. I received this
proof on May 28, 1940.

b. These four proofs show high Intellectual
ability, and prove what boys and girls can do when
permitted to think Independently and logically.

E. S. Loomis.

c. This proof belongs in the book at the end
of the E-Type section. One Hundred Twenty-Six.

GEOMETRIC PROOFS

265

Iwft.llu.adLed-ELfti-SL?.

Geometric proofs are either Euclidian, as the
preceding 255, or Non-Euclldlan which are either
Lohachevsklan (hypothesis, hyperbolic, and curvature,
negative) or Rlemanlan (hypothesis. Elliptic, and
curvature , positive ) .

The following non-euclldlan proof Is a liter-
al transcription of the one given In "The Elements
of Non-Euclldlan Geometry," (1909), by Julian Lowell
Coolldge, Ph.D., of Harvard University. It appears
on pp. 55-57 of said work. It presumes a surface of
constant negative curvature, --a pseudo sphere, --hence
Lohachevsklan; and Its establishment at said pages
was necessary as a "sufficient basis for trigonome-
try," whose figures must ^.ppear on such a surface.

The complete exhibit In said work reads:

"Let us not fall to notice that since Is

a right angle we have, (Chap. Ill, Theorem 17),

RP Tt

llm. = cos (— - 0 ) = sin 0. ( 3 )

"The extension of these functions to angles

whose measures are greater than will afford no

difficulty, for, on the one hand, the defining series
remains convergent, and, on the other, the geometric
extension may be effected as In the elementary books.

"Our next task Is a most serious and funda-
mental one, to find the relations which connect the
measures and sides and angles of a right triangle.

Let this be the AABC with ^ ABC as the right angle.

Let the measure of ^-BAC be while that of ^BCA Is 0.

We shall assume that both and 9 are less than ,

an obvious necessity under the euclidian or hyper-
bolic hypothesis, while under the elliptic, such will
still be the case If the sides of the triangle be not
large, and the case where the Inequalities do not
hold may be easily treated from the case where they
do . Let us also call a, b, c the measures of BC, CA,
AB respectively.

266

THE PYTHAGOREAN PROPOSITION

”We now make rather an elaborate construction.
Take In (AB) as near to B as desired, and Ai on

the extension of (AB)
beyond A, so that AAi
= BBi and construct
AAiBiCi = ABC, Cl lying
not far from C; a con-
struction which, by 1
(Chap. IV, Theorem 1 ),

Is easily possible If
BBi b e small enough. Let
BiCi meet (AC) at C^.
^CiC2C will differ but

little from ^CA, and we

^ may dra w Ci Ca perpendic-

ular to CCa, where C3 Is
Fig. 2 a point of (CC2). Let

us ne xt f ind A2 on the
extension of (AC ) beyond A so that A^ = 0 ^ and B2
o n the extension of (CiBi ) beyond Bi s o tha t B1B2
“ C1C2, which Is certainly possible as C1C2 Is very
small. Draw A2B2 . We saw that ^.CiC2C will differ
f rom 4 -BCA by an Infinitesimal (as BiB decreases) and
J.CCiBi will approach a right angle as a limit. We
thus get two appr oxlma t e expressions of sin 0 whose

comparison yields + e. = QQS BBi ^ ^

C1C 2 CC2 CC2

fo r CC i - cos a/k BBi Is Infinitesimal In comparison
to BBi or CCi . Again, we see that a line through the
middle point of (AAi ) pe rpendicular to AA2 will also
be perpendicular to AiCi, and the distance of the In-
tersections will differ Infinitesimally from sin tfAAT.
We see that C1C2 differs by a higher Infin ites imal

from sin cos b/k AAi, so that cos 7- slni|>'=-+ e-
cos a/k Bil . CiCa

— ' '■ —aa— — +

CCi ^

"Next we see th at AAi = BBi, and hence cos -
1 C1O2 ^

sin T|» ^ + ® 4 • Moreover, by

JAMES JOSEPH SYLVESTER
1814-1897

SOME NOTED PROOFS

Of the 370 demonstrations, for:

Proof

1. The shortest, see p. 24, Legendre* s One

2. The longest, see p. 8I, Davies Legendre .. Ninety

3. The most popular, p. 109, Sixteen

4. Arabic, see p. 121; under proof .... Thirty-Three

5. Bhaskara, the Hindu, p. 50, Thirty-Six

6. The blind girl, Coolidge, p. II8, .... Thirty-Two

7. The Chinese — before 500 B.C., p. 26l,

Two Hundred Fifty-Three

8. Ann Condlt, at age I6, p. l40 (Unique)

Sixty-Eight

9. Euclid's, p. 119, Thirty-Three

10. Garfield's (Ex-Pres. ), p. 231,

Two Hundred Thirty-One

11. Huygens* (b. 1629), p. II8, Thirty-One

12. Jashemskl's (age I8), p. 230,

Two Hundred Thirty

13. Law of Dissection, p. IO5, Ten

14. Leibniz's (b. 1646), p. 59, Fifty-Three

15. Non-Euclldlan, p. 265, ... Two Hundred Fifty-Six

16. Pentagon, pp. 92 and 238,

. One Hundred Seven and Two Hundred Forty-Two

17* Reductlo ad Absurdum, pp. 4l and 48,

Sixteen and Thirty-Two

18. Theory of Limits, p. 86, Ninety-Eight

268

ADDENDA

269

They came to me from everywhere .

1. In 1927 , at the date of the printing of the 1st
edition, it shows — No. of Proofs:

Algebraic, 58 ; Geometric, l67; Quater-
nlonlc, 4; Dynamic, 1; in all 230 dif-
ferent proofs.

2. On November I 6 , 1933, niy manuscript for a second
edition gave:

Algebraic, 101; Geometric, 211; Quater-
nlonlc, 4; Dynamic, 2; in all 318 dif-
ferent proofs.

3 . On May 1, 1940 at the revised completion of the
manuscript for my 2nd edition of The Pythagorean
Proposition, it contains — proofs:

Algebraic, 109; Geometric, 255; Quater-
nlonlc, 4; Dynamic, 2; in all 370 dif-
ferent proofs, each proof calling for
its own specific figure. And the end
is not ye*^ .

E. S. Loomis, Ph.D.
at age nearly 88,
May 1, 1940

BIBLIOGRAPHY

The sources, from which historical facts,
suggestions, illustrations, demonstrations and other
data concerning the preparation of this book, or for
further investigation by any one who may wish to com-
pare and consider original sources, are comprised In
the following tabulation.

I. TEXTS ON GEOMETRY

1. Beman and Smithes Plane and Solid Geometry, 1895.

2. Bezout*s Elements of Geometry, I 768 .

3 . Gamer er, J. G. et C. P. Hanbes-Berel, 1824.

4. Chauvenet*s Elements of Geometry, I 887 .

5 . Cramer, (C.), 1837, 93 proofs.

6. Davies* Legendre, I 858 .

7 . Dobrlner*s (Dr. H. ) Geometry.

8. Edwards* Geometry, 1895.

9 . Halste.d*s Elementary Greometry, l895.

10. Hlll*s Geometry for Beginners, I 886 .

11. Hopkins* Plane Geometry, I 891 .

12. Kunze*s (van) Geometry, 1842.

13 . Mackay*s (J. S. ) Plane Geometry.

14. Mahler *s (G. ) "Ebene Geometrle,” 1897.

15 . Milne *3 Plane and Solid Geometry, 1899.

16 . 01ney*s Elements of Geometry, 1872, Unlv*y Ed*n.

17 . Pisano (Leonardo) Practica Georaetrlae.

18 . Ray*s Geometry.

19 . Richards* (Claudius) "Euclldes Elementoriam
Geometricorum, " 1645, Antwerplae .

20. Sauvens* (M. ) Geometry, 1753.

21. Schuyler *s (Dr. A.) Elements df Geometry, I 876 .

22. Simpson* s (Thos. ) Elementary Geometry.

23 . Todhunter*s Elements of Euclid, 1897.

24. Wells* Essentials of Geometry, 1899.

25 . Wentworth and Smith* s Plane Geometry, Revised Edi-
tion, 1910 .

26 . Wentworth* s New Plane and Solid Geometry, 1895.

271

272

THE PYTHAGOREAN PROPOSITION

II. OTHER TEXTS

1. Bernstein, (P. ) The Axiomatic Simplicity of
Proofs, 1909.

2. Blanchard* s (Orlando) Arithmetic.

3. Brandes* Dissection Proofs, 1908.

4. Casey* s Sequel to Euclid, Part I, I9OO.

5. Cours de Mathematlques, 1768.

6. Cullen* 3 (P. C.) Pamphlet.

7. Delboeuf*s work.

8. Pourrey*s (E. ) Curiosities Geometrlques, 2nd
Ed*n, 430 pages, 1778, Paris.

9. Geometric Exercises, Paper Folding, 1905.

10. Halsted*3 Mensuration.

11. Handbook der Mathematlk, I869.

12. Hardy* 3 Elements of Quaternions, I88I.

13. Hubert *s Rudlmenta Algebrae, 1792.

14. Leltzinann*s (Dr. W. ) "Des Pythagorlsche Lehrsatz,”

1930.

15. Llttrow*3 (E, von) "Popular Geometrle," 1839.

16. Loomis (Dr. E. S. ), The Pythagorean Proposition.

17. Row*s Geometric Exercises In Paper Folding, 1905.

18. Schubert *3 Mathematical Essays, I898.

19. Schuyler* 3 (Dr. A.), Trigonometry, l873.

20. Simon (M. ), I906.

21. Vogt (H. ), The Geometry of Pythagoras, Vol. 3#

1909.

22. Wallis* Treatise of Algebra, I685.

III. MONOGRAPHS

1. Heath* s Mathematical Monographs, Nos. 1 and 2,

1900.

2. Hoffmann* s (j. J. I,), The Pythagorean Theorem,
1821—32 Proofs.

3. Martin’s (Artemus), "On Rational Right-Angled
Triangles (Pamphlet), 1912.

4. Mathematical Monograph, No. I6, 1915, Dlaphontlne
Analysis des Pythagoralsche Lehrsatz, by Leltz-
mann.

BIBLIOGRA.PHY

273

5. Naber (Dr. H. A. ), Das Theorem des Pythagoras,

1908 .

6 . Versluys* (J. ), Theorem van Pythagoras, Zes en
Negentlg Bewljzen, 191^^ 96 Proofs.

7 . "Vriend de Wlskunde,” I 898 by Vaes (P, J.).

8 . Wlpper, (Jury), ”46 Bewelse des Pythagoralschen
Lehrsates,” I 88 O, 46 Proofs.

IV. JOURNALS AND MAGAZINES

1. American Mathematical Monthly, 1894-1901, Vols. I
to VIII Inclusive, 100 proofs. The Editor, Dr.

B. P. Yanney said: ”There Is no limit to the
number of proofs — we just had to quit.”

2. Archlv der Mathematlk and Phys., Vols. II, XVII,
XX, XXIV, 1855 , by Ph. Grunert, l857.

5 . Aumerkungen uber Hrn. geh. R, Wolf*s Ausgug aus
der Geometric,” 17^6, by Kruger.

4. Bernstein (P. ), The Pythagorean Theorem, 1924.

5 . Companion. to Elementary School Mathematics, 1924.

6. Heutton*s (Dr.), Tracts — History of Algebra, 5
Vols., 1912 .

7 . Journal of Education, 1885-8, Vols. XXV, XXVI,
XXVII, XXVIII.

8. Journal de Matheln, I 888 , by P. Pabre.

9 . Journal of the Royal Society of Canada, 1904,

Vol. II, p. 239 .

10. Lehrbegrlf f » s Science of Mathematlk.

11. Masonic Bulletin, Grand Lodge of Iowa, No. 2,
Pebruary 1929-

12. Mathematical Essays and Recreations.

13 . Mathematical Magazine, The, Vols. I and II, by
Martln> I 89 I.

14. Mathematical Monthly, Vols. I and II, I 858 , 28
Proofs, by J. D. Runkle.

15 . Mathematics Teacher, Vol. XVI, 1915, Vol. XVIII,

1925 .

16 . Messenger of Mathematics, Vol. 2, I 873 .

17 . Nengebrauer * s (O. ), History of the Pythagorean
Theorem, I 928 .

27k THE PYTHAGOREAN PROPOSITION

18. Notes and Queries, by W. D. Henkle, l875-8l.

19. Nouvelles Annales, Vol. LXII.

20. Open Court Pub. Co., I898, p. 59 — Magic Squares.

21. Periodlco dl Mathematlcke (Italian), History of
Che Pythagorean Theorem.

22. Phllosophia et Ma thesis Unlversa.

25. Philosophical Transactions, I685.

24. Pythagorean Theorem, The, by Cullen.

25. Quarterly Joiirnal of Mathematics, Vol. I.

26. Recreations In Mathematics and Physics, 1778.

27. Runkle’s Mathematical Monthly, No. 11, 1859-

28. School Visitor, The, Vol. Ill, 1882, Vol. IX,

1888.

29. Schbrer*s The Theorem of Pythagoras, 1929.

50. Science, New Series, Vol. 52, 1910; Vol. 53, 1911;
Vol. 54, 1912.

51. Sclentla Baccalurem.

52. The Scientific American Supplement, Vol. 70,

1910, pp. 559, 382-5.

55. Sclentiflque Revue, 1859.

V. GENERAL REFERENCE WORKS

1. Ball’s Short History of Mathematics, I888.

2. The Encyclopedia Brltannlca — articles on Geometry,
Magic Squares, Dynamics, Quaternions or Vector
Analysis, Euclid, Pythagoras, etc.

5. Encyclopadle, des Elementar Mathematlck, 1925,
by H. Weber and J. Wellstein.

4. Scanscrlt Mathematics.

(1) ”Vlga Ganlta.”

(2) "Yoncti Bacha.”

5. Wolf’s (Dr. Rudolph), Handbook of Mathematics, I869

6. MUller (J. W. ), Nlirnberg, I819.

BIBLIOGRAPHY

275

VI.

The Royal Society of London- -Index 1800-1900;
Vol. I, Pure Mathematics, 1908, lists the following
authors and articles:

1. Muller, E. K. F. von: Oken Isis (1826), 763ff.

2. Gundermann, C. : Crelle, J. 42 (1852), 280ff.

3. Vincent, A. J. M.: N, A. Mth. 11 (1852), 5ff.

4. DeMorian, A.: Q, j. Mth. 1 (l857), 236ff.

5. Sal wen, G. : Q. j. Mth. 1 (1857), 237ff.

6. Slchardson, J. M.: Camb. (M. ) Math. M. 2 (i860),
45ff.

7. Azzarelli, M. 1873 : Rm. N. Line. At. 27 (18741
66ff.

8. Harvey, V.: Edlnb. Math. S. P. 4 (l886), 17ff.

9. Zahradnlk, JT. : Casopls 25 (1896), 26lff. Schr.
Mth. (1896), 364ff .

10. Confirmation. Vlttsteln, T.t Grxmert Arch. 11
(1848), 152.

11. Pythag. triangle. Grunert, J. A.: Arch. der.
Mth. and Physlk 31 (I858), 472ff.

12. Demonstration. Boquoy, G. von: Oken Isis 21

(1828), Iff.

13. by dissection. Periial, B.: Mess.

Mth. 2 (1873), 103ff.

14. Extension to any triangle. Jelinek, K.; Casopls
28 (1899), 79ff.. P. Schr. Mth. (l899), 456.

15. Extensions. Anglin, A. B,: Edlnb. R. S. P. 12

(1884) 703.

16. Figures for proving. Andre, H. d': N. A. Mth.

5 (1846), 324.

17. Generalization. Umpfenbach (Dr.): Crelle J. 26
(1843), 92.

18. Puzzles connected with theorem. Brand, B.: Rv.
Sc. 2 (1894), 274ff., 809ff.

19. Pythagorean triangles. Cantor, M.: Schlonsllch,

Z. 4 (1859), 306ff.

20. Pythagorean triangles. Lewie, J.: Am. Ph. S. P.
9 (1865), 4l5ff.

21. Pythagorean triangles. Vhitworth, V. A.: Lpool,

Ph. S. P. 29 (1875), 237ff.

276

THE PYTHAGOREAN PROPOSITION

22. Pythagorean triangles and applications to the
division of the clrcijmference. Grueben Arch.
Mth. und Physlk 15 (1897), 337ff.

25. Pythagorean triangle, nearly Isosceles. Martin,
Art: Des Moines Anal ? (1876), 47ff.

24. Pythagorean triangle, properties. Breton: Ph.
Les Monde 3 6 (l864), 401ff.

25. Right-angled triangles. Banima, V, S. : Amst.

Vh. 4 (1818), 28ff.

26. Right-angled triangles, squares on sides (Harnett).
Gerionne, J. D.i Gergonne A. Mth. l4 (1825),
534ff .

27. Quadrangles, complete. Mewberi, J. : Mathses 11

(1891), 33ff, 67ff, 8lff, l89ff.

28. Quadrangles, transformations. Gob, A.: As. Pr.
C. R. (1884), (Pt. 2), 294ff.

TESTIMONIALS

Prom letters of appreciation and printed Re-
views the following four testify as to Its worth.

New Books. The Mathematics Teacher, 1928,
has: The Pythagorean Theorem, Elisha S. Loomis,

1927, Cleveland, Ohio, 2l4 pp. Price $2.00.

"One hundred sixty-seven geometric proofs
and fifty-eight algebraic proofs besides several
other kinds of proofs for the Pythagorean Theorem
compiled In detailed, authoritative, well-organized
form will be a rare 'find* for Geometry teachers who
are alive to the possibilities of their subject and
for mathematics clubs that are looking for Interest-
ing material. Dr. Loomis has done a scholarly piece
of work In collecting and arranging In such conven-
ient form this great number of proofs of our historic
theorem.

"The book however Is more than a mere cata-
loguing of proofs, valuable as that may be, but pre-
sents an organized suggestion for many more original
proofs. The object of the treatise Is twofold, *to
present to the future Investigator, under one cover,
simply and concisely, what Is known relative to the
Pythagorean proposition, and to set forth certain es-
tablished facts concerning the proofs and geometric
figures pertaining thereto.*

"There are four kinds of proofs, (l) those
based upon linear relations — the algebraic proof,

(2) those based upon comparison of areas — the geo-
metric proofs, (5) those based upon vector operations
— the quaternlonlc proofs, (4) those based upon mass
and velocity — the dynamic proofs. Dr, Loomis con-
tends that the number of algebraic and geometric
proofs are each limitless, but that no proof by trig-
onometry, analytics or calculus Is possible due to

277

278

THE PYTHAGOREAN PROPOSITION

the fact that these subjects are based upon the rlght-
trlangle proposition.

"This book is a treasure chest for any mathe-
matics teacher. The twenty-seven years which Dr.
Loomis has played with this theorem is one of his
hobbies, while he was Head of the Mathematics Depart-
ment of West High School, Cleveland, Ohio, have been
well spent since he has gleaned such treasures from
the archives. It is impossible in a short review to
do justice to this splendid bit of research work so
unselfishly done for the love of mathematics. This
book should be highly prized by every mathematics
teacher and should find a prominent place in every
school and public library."

H. C. Chrlstoffenson

Teachers College

Columbia University, N.Y. City

Prom another review this appears:

"It (this work) presents all that the litera-
ture of 2400 years gives relative to the historically
renowned and mathematically fundamental Pythagorean
proposition — the proposition on which rests the sci-
ences of civil engineering, navigation and astronomy,
and to which Dr. Einstein conformed in formulating
and positing his general theory of relativity in

1915.

"It establishes that but four kinds of proofs
are possible — the Algebraic, the Geometric the Quater-
nionlc and the Dynamic.

"It shows that the number of Algebraic proofs
is limitless.

"It depicts 58 algebraic and I 67 geometric

proofs.

"It declares that no trigonometric, analytic
geometry, or calculus proof is possible.

"It contains 250 geometric figures for each
of which a demonstration is given.

TESTIMONIALS

279

"It contains a complete bibliography of all
references to this celebrated theorem.

"And lastly this work of Dr. Loomis Is so
complete In Its mathematical survey and analysis that
It Is destined to become the reference book of all
future Investigators, and to this end Its sponsors
are sending a complimentary copy to each of the great
mathematical libraries of the United States and
Europe . "

Masters and Wardens Association
of the

22nd Masonic District of Ohio

Dr. Oscar Lee Dusthelmer, Prof, of Mathe-
matics and Astronomy In Baldwln-Wallace College,
Berea, Ohio, under date of December 17, 1927, wrote:
"Dr. Loomis, I consider this book a real contribution
to Mathematical Literature and one that you can be
justly proud of.... I am more than pleased with the
book. "

Oscar L. Dusthelmer

Dr. H. A. Naber, of Baarn, Holland, In a
weekly paper for secondary Instructors, printed, 193^,
In Holland Dutch, has (as translated): "The Pytha-
gorean Proposition, by Elisha S. Loomis, Professor
Emeritus of Mathematics, Baldwln-Wallace College,"
(Bera, 0. )

Dr. Naber states .... "The author has classi-
fied his (257) proofs In groups: algebraic, geometric,
quaternlonlc and dynamic proofs; and these groups are
further subdivided?" "....Prof. Loomis himself has
wrought. In his book, a work that Is more durable
than bronze and that tower higher even than the pyra-
mids." "....Let us hope — until we know more com-
pletely- -that by this procedure, as our mentality
grows deeper ^ It will become as In him: The Philo-
sophic Insight."

INDEX OF NAMES

1, Names of all authors of works referred to or consulted in
the preparation of this hook may he found In the hlhllogra
phy on pp. 271 - 76 .

2 . Names of Texts^ Journals^ Magazines and other publications
consulted or referred to also appear In said hlhllography.

3* Names of persons for whom a proof has heen named^ or to
whom a proof has heen credited, or from or through whom a
proof has come, as well as authors of works consulted, are
arranged alphabetically In this Index of personal names,

4. Some names occur two or more times, hut the earliest occur
rence is the only one paged.

Adams, J,, 60
Agassiz, 190
Amasls, 8
Anakslmander, 8
Andre, H. d», 98
Anglin, A. H., 275
Andronlcus, 15
Annalrlzo of Arabia,

900 N.C,, 51
Arabic proof, 121
Ash, M. V. ( 1683 ), 150
Azzarelll, M., 275

Ball, W. W. House, 154
Bangma, V. S., 276
Bauer, L. A., 79
Bell, E. T, (Dr.), vl
Bell, Hichard A., 49
Bernstein, F., 272
Bhaskara, the Hindu, 50
Blnford, R. E., 230
Blanchard, Orlando, 155
Boad, Henry, 1733, 137

Boon, F, C., 66
Boquoy, G. von, 275
Bottcher, J. E,, 112
Brand, E., 65
Brandes, 272
Bressert, M. Plton, 67
Breton, 276

Brets chenschneider, 155
Brown, Lucius, 79
Buck, Jenny de, l49
Bullard, L. J., 83

Calderhead, Prof. James A.,

247

Camerer, I 3
Camirs, Jules, 52
Cantor, M., 5
Carmichael, Robert D., 24
Casey, 253
Cass, Gustav, 226
Cattell, J. McK., 249
Chase, Pliny Earle, IO 9
Chauvenet, 52

281

282

THE PYTHAGOREAN PROPOSITION

Chlllag, Bob, 263
Chinese proof, 262
Chrlstof Person, H. C,, 278
Clalraut, 203

CollDurn, Arthiir E., 53
Co Ulna, Matthew, I8
Condlt, Aim (a^e I6), li^-0
Coolid^e, Miss E. A.

(■blind girl), 119
Coolldge, Julian Lowell, 265
Copernicus, 12
Cramer, C., 27I
Crelle, J., 275
Cullen, R. C., 155

Darwin, 121a
Davies (Legendre), 24
DeBeck, Bodo M., 227
Deimal, R. F., 250
Deltoeuf, 1^4
DeMorgan, A,, 105
Descartes, 244
Dickinson, M., 179
Dickson, Leonard E., I7
Diogenes, 6

Dissection, Law of — Loomis, 106
Dotrlner, Dr. H., Ill
Dulfer, A. E. B., 57
Dusthelmer, Dr. Oscar L., 279

Edison, 12
Edwards, 25
England, W. , 120
Epstein, Paul, 115
Erat Okies, 9
Euclid, 13
Evans, Geo, W., 57
Excell, Rev, J, G, , 49

Ferekld, 9
Fortes, E,, 203
Fourrey, E., 49
French, C., 133

Galileo, 12

Garfield, Pres, James A., 224
Gauss, 16

Gelder, Jacob de, 121
Gergonne, J, D., 276
Glnsburg, Dr, Jehuthlel, xlv
Glashler, J. W. L., 17
Got, A., 276
Graap, F., 11
Grueher, 276
Grundermann, C,, 275
Grunert, J, A,, 53
Guthell, B, von--W6rld War
Proof, 117

Halsted, Geo, B,, 21
Hamlet, 120
Hardy, 272
Harvey, W., 275
Hauff*, von, 123
Hawkins, Cecil, 57
Haynes, 226
Heath, 24
Henkle, W. D., 274
Hersey, Mayo D,, 249
Hill, 271
Hlpplas, 9

Hofftnann, Joh, J, I., I818,

29

Hoghen, Lancelot, 261
Hoover, Wm., , I55
Hopkins, G, I,, 68
Horace, 285
Houston, Joseph, 145
Hoyt, David W., IO9

Fatre, F., 57

INDEX OP NAMES

28 ?

Hubert!, I762, 225
Hutton, Dr,, 48
Huygens, L657, II8
Hypasos, 11
Hyps idem, I57

Ingraham, Andrew, 245
Isidorum, I56

Jackson, C. S., 226
Jashemskl, Stanley, 84
Jelinek, v,, 232
Jengis, Khan, I55
Joan, R,, 177

Johnston, Theodore H., xlil
Jowett, 86

Eamhls, 9
Kemper, C, J., 70
Khayy^, Omar, 120a
Klagel, 151
Khoer, Alvin, 4o
Erfiger, M. , 110
Krueger, 1746, 1
Kruithosch, D. J., I39
Kunze, von, 192

Laertius, 6
Laisnez, Maurice, 50
Lamy, R. P., I685, 85
Lecchio, 125
Legendre, Andren M,, 24
Lehman, Prof. D, A,, 4o
Leihniz, von, 59
Leonardo da Vinci, 129
Lewis, J., 275
Leltzmann, Dr. Wm., 49
Littrow, E. von, II5
Locke, J. B., 248
Loomis, Elatus G., xlv
Loomis, Dr. E. S., I58

Macay, 271

MacFarlane, Alexander, 249
MaJiler, von G., I05
Marconi, 12
Marre, A., 50
Martin, Artemaa, 17
Martin, Fred W., 263
Masares, 20
McCready, J. M., 174
McIntosh, M., 209
Meyer, P. Armand, 44
Milne, 271
Mhessarch, 7
M6 liman, E., 79
Muller, J. W., 274

Naher, Dr. H. A., 275
Nasir-Ed-Dln, Persian Astron-
omer, 128
Nengebrauer, 273
Newberg, J., 276
Newton, Sir Isaac, 12
Nielson, Chr., Il4
Northrup, Edwin F., 249

Oliver, 83
Olney, 73

Ozanam, M. de C. G., I778,

110

Pappus, a. 375 A.D., 126
Perlgal, Henry, 102
Philippi, III of Spain, I56
Phillips, Dr. Geo. M., I83
Pisano, Leonardo, 52
Pit hay, 7

Plton-Bressant, 67
Plato, 17
Plutarch, 6
Poly crates, 7

284

THE PYTHAGOREAN PROPOSITION

PosthumuB^ J. J.^ 92
Proolos, 4
Psammenlt, 9
Ptolemus (Ptolemy), 66
Pythagoras, 6

Raub, M,, 66

Ray, Prof. Saradarujan (in-
dla), 155

Relchenherger, Nepomucen, 177
Renan, H., 46
Richards, Claudius, I56
Richardson, Prof. John M., 26
Rogot, M., 180
Row, T. Sundra, 100
Runkle, J. D«, 66
Rupert, Wm. W. , 23

Salwen, G., 275
Sarvonarola, 12
Saunderson, 229
Saureur, 227
Schau-Gao, 5
Schiamllch, 194
Schooten, van, 203
Schorer, 274
Schubert, Herman, 256
Schuyler, Hr. Aaron, 64
Simon, 272

Simpson, David P., xlll
Slinpson, Thomas, I55
Skats chkow, 5
Smedley, Pred S., 90
Smith, David E., 271
Smith, Dr. Wm. B., I55
Socrates, 86
Sonchls, 9

Sterkenberg, C. G., 40
Stowell, T. P., 224
Sturm, J. C., 203
Sunderland, I58

Tarquln, 9

Templehoff, von, I769., 129
Thales, 8
Theana, 10

Thompson, J. G., 234
Thornton, Wilson, 50
Todhunter, Dr., 27I
Towne, Paul W., 255
Trowbridge, David, I52
Ts Chou -Gun, 5
Tyron, C. W., 110

Umpfenbach, Dr., 275
Uwan, 5

Vaes, F. J., 46
Versluys, J., 13
Vleth, van, 132
Vincent, A. J. N., 275
Vinci, Leonardo da, 129
Vogt, H., 272
Vulbert, de, 47

Wallis, John, 52
Warlus, Peter, l48
Waterhouse, John, 252
Weber, H., kO
Wells, 271
Wellsteln, T., 40
Wentworth, 52
Werner, Oscar, 53
Wheeler, Rev. D. A., 49
Whitworth, 275
Wlpper, Jury, 3
Wolf, Rudolph, 274

Yanney, Prof. B. F., 242
Young, Dr. Charles A., 70

Zahradnlk, 275
Zelson, Joseph, 122
Zelson, Prof. Louis G., l4l

”^Exeii nonunentum, aere perennius
Beiali que situ pyramidum altius.

Quod non imber edax, non aquilo impoteno
Possit diruere aut Innumerabilis
Annorun series et fuia temporuni.

Hon omnis mortar. "

— Horace
30 ode in
Book III